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Full text of "Geometry"




WILEY 
BENZIGER 



GEOMETRY 

UNDER THE EDITORIAL DIRECTION OF 

ROY DUBISCH AND ISABELLE P. RUCKER 



LAWRENCE A. RINGENBERG 
RICHARD S. PRESSER 



Benziger, Inc. New York, Beverly Hills 

'',n Association v/itn 

John Wiley & Sons, Inc. New York, London, Sydney, Toronto 



EDITORS 



ROY DUBISCH, Professor of Mathematics at the University of Washington, has 
been writing for both teachers and students for many years. He has been Associate 
Editor of the American Mathematical Monthly and Editor of the Mathematics Maga- 
zine. He has several books to his credit including The Teaching of Mathematics 
(Wiley, 1963) and is the author of numerous journal articles. Since 1961 Profes- 
sor Dubisoh has served on the Mathematics Steering Committee of the African Edu- 
cation Project and has participated in several workshops and institutes in various 
African countries. He has also directed and lectured at several NSF summer in- 
stitutes. He is a past Vice President of the National Council of Teachers of 
Mathematics and has served on the Teacher Training Panel of the Committee on the 
Undergraduate Program in Mathematics, the Advisory Board and Panel on Supple- 
mentary Publications of the School Mathematics Study Group, and the Board of 
Governors of the Mathematical Association of America. He has also been active in 
many other professional organizations, 

ISABELLE P. RUCKER, Supervisor of Mathematics of the State Board of Education 
of Virginia, is a former teacher of both elementary and secondary school mathe- 
matics. She has participated in summer mathematics institutes and an NSF 
Academic Year Institute at he University of Virginia and has published numerous 
articles and book reviews in professional journals. In addition to being active in 
other professional organizations, Mrs. Rucker has served on the Advisory Board, 
the Executive Committee, and the Panel on Supplementary Publications of the 
School Mathematics Study Group, For the National Council of Teachers of Math- 
ematics, she has served as Chairman of the Committee on Plans and Proposals 
and as a member of the Committee on Affiliated Groups, She was Program Chair- 
man for the Golden Jubilee Annual Meeting in 1970 and Is an active member of 
the Committee on Meetings of the Council. In addition, she has held offices in the 
Association of State Supervisors of Mathematics, 



Copyright © 1973 by John Witty and $om, Inc. 

All nghu r*t»rv*d, PubJfthctf «lmullan»t>im|y )n Canoda. 

No port of ihl» book may be r»pr«doe»d by any mtoni, nor 
frflnimi««d, nor troniloted inlo a moctoln* lannyoflfl 
wifhoul Iho wriltfln p«rmin1on of (h» pvblUhir, 

ISBN 0-471 -P4549.8 

Printed In ihe UnFfsd Sratoi of Afntrka 

10 967*34321 



AUTHORS 



RICHARD 5. PRESSER, Coordinator of Secondary School Mathematics of the 
Michigan City Schools, Michigan City, Indiana, is a former writer for the Minne- 
sota National Laboratory, the School Mathematics Study Group, and the African 
Education Program of Educational Services, Inc. Mr, Presser has many years of 
teaching experience at the secondary school level and is active In the Indiana 
Council of Teachers of Mathematics and the Central Association of Science and 
Mathematics Teachers. 

LAWRENCE A. RINGENBERG, Professor of Mathematics and Dean. College 
of Letters and Science, Eastern Illinois University, Charleston, Illinois, was Head 
of the Department of Mathematics from 1947 to J 967. He is a former writer for 
the School Mathematics Study Group, the author of College Geometry (Wiley, 
1968) and Informal Geometry (Wiley, 1967), and is currentry serving as a mem- 
ber of the editorial board and one of the authors of the forthcoming NCTM Year- 
book on geometry. Dean Ringenberg was Coordinator of the U.S.A.LD. Mathe- 
matics and Physics Courses Program in East Pakistan for Intermediate College 
teachers in East Pakistan during the summer of 1966. Since 1960 he has been 
a member of the Charleston, Illinois. Unit District Board of Education, 



Preface 



Geometry is one of a series of five mathematics textbooks for 
junior and senior high schools. It is designed for a one-year course 
for students with a background of informal geometry and elementary 
algebra such as that included in Mathematics J, Mathematics II, and 
Algebra of this series. 

In its formal development of Euclidean geometry, this textbook 
features an integrated treatment of plane and solid geometry with 
an early introduction of coordinates. Coordinates on a line, in a 
plane, and in space relate numbers ami points. They are used to 
gain knowledge of geometrical figures and to simplify the develop- 
ment of formal geometry, 

Students use their knowledge of elementary algebra throughout 
the book to help them learn geometry. In doing this they maintain 
and strengthen their competence in algebra. 

The postulates in this book form the basis of a rigorous, yet 
plausible, development of a first course in formal Euclidean geometry. 
In some instances, statements that are proved in more advanced 
treatments are accepted as postulates here. This has been done to 
decrease the length of the development and Lo make the develop- 
ment appropriate for high school students. 

Geomctrv is an important subject because it is practical and 
useful and at the same lime abstract and theoretical. There arc two 
main objectives in this geometry textbook. One is to help students 
learn a body of important facts about geometrical figures. These 
facts, interpret* all v. are facts about the space in which we 



will Preface 



live. These facts are important for intelligent citizenship and for suc- 
cess in many careers. The other main objective is to help students 
attain a degree of mathematical maturity. In the elementary and 
junior high schools, students are encouraged to learn by observing 
and manipulating physical objects. There is considerable emphasis 
On intuitive and inductive reasoning. The power and beauty of 
mathematics, however, is due primarily to its abstractness. The 
generality of its theorems makes possible a variety of applications. 
Understanding mathematics "*in die abstract" is tantamount to 
understanding the deductive method in mathematics. In studying 
this book students will develop their capacity to reason deductively 
and hence their ability to read and write proofs. 



Contents 



Chapter 1 
Points, Lines, and Planes 1 

I.l Informal Geometry 1 / 1.2 The Idea of Formal 
Geometry 4 / 1.3 The Ideas of Point, line, and Plane 
10 / I A Sets 14 / 1.5 Conjunctions and Disjunctions 
18 / 1.6 The Incidence Relationships of Points and Lines 
22 / 1.7 The Incidence Relationships of Points, T.incs, and 
Planes 28 / Chapter Summary 34 / Review Exercises 35 

Chapter 2 

Separation and Related Concepts 40 

2.1 Introduction 41 / 2.2 The Betweenness Postulates 
41 / 2.3 Using Betweenness to Make Definitions 47 / 2.4 
The Concept of an Angle 52 / 2.5 The Separation Postulates 
56 / 2.6 Interiors and Exteriors of Angles 67 / 2.7 
Triangles and Quadrilaterals 72 / 2.8 Properties of Equality 
and Number Operations 80 / 2*9 Solving Equations 
S3 / 2.10 Equivalent Equations 86 / Chapter Summary 
92 / Review Exercises 93 

Chapter 3 

Distance and Coordinate Systems 96 

3.1 Introduction 97 / 3.2 Distance 98 / 3.3 Line 
Coordinate Systems 105 / 3.4 Rays, Segments, and 
Coordinates 107 / 3.5 Segments and Congruence 



Contents 

111 / 3.6 Two Coordinate Systems on a Line 116 / 3.7 
Points of Division 129 / Chapter Summary 135 / Review 
Exercises 135 

Chapter 4 

Angles, Ray-Coordinates, and Polygons 138 

4.1 Introduction 139 / 4.2 Angle Measure and 

Congruence 140 / 4.3 Betweenness for Rays 146 / 4.4 

Ray-Coordinates and the Protractor Postulate 152 / 4*5 

Some Properties of Angles 160 / 4,6 Interiors of Angles 

168 / 4.7 Adjacent Angles and Perpendicularity 

170 / 4,8 Polygons 177 / 4,9 Dihedral Angles 

181 / Chapter Sunn nary 184 / Review Exercises 185 

Chapter 5 
Congruence of Triangles 188 

5.1 Introduction 189 / 5.2 Congruence of Triangles 
190 / 5.3 "If-Thcn" Statements and Their Converses 
197 / 5.4 The Use of Conditional Statements in Proofs 
200 / 5.5 Proofs in Two-Column Form 203 / 5M The 
Congruence Postulates for Triangles 208 / 5.7 Using the 
S.A.S., A.S.A.. and S.S.S. Postulates in Wilting Proofs 
216 , 5.8 Isosceles Triangles 227 / 5.9 Medians and 
Perpendicular Bisectors 234 / Chapter Summary 242 / 
Review Exercises 244 

Chapter 6 

Inequalities in Triangles 246 

6.1 Introduction 247 / 0,2 Inequalities for Numbers 
248 / 6.3 The Exterior Angle Theorem 254/6.4 
Inequalities Involving Triangles 262 / Chapter Summary 
274 / Review^ Exercises 275 

Chapter 7 
Parallelism 280 

7.1 Introduction 281 / 7.2 Definitions 282 / 7.3 
Existence of Parallel Lines 284 / 7.4 Transversals and 
Associated Angles 286 / 7.5 Some Parallel Line Theorems 



Contents xl 



/ 7.0 The Parallel Postulate and Some Theorems 
301 / 7.7 Parallelism for Segments; Parallelograms 
308/7.8 Parallelism for Rays 314 / 7.9 Some Theorems 
on Triangles and Quadrilaterals 318 / Chapter Summary 
325 / Review Exercises 325 

Chapter 8 
Perpendicularity and Parallelism in Space 328 

8.1 Introduction 329 / 8.2 A Peipendieularity Definition 
332 / &3 A Basic Perpendicularity Theorem ,334 / 8.4 
Other Perpendicularity Theorems 333 / 8.5 Parallelism for 
Lines and Planes 347 / 8.6 Parallelism and Perpendicularity 
352 / Chapter Summary 359 / Review Exercises 380 

Chapter 9 
Area and the Pythagorean Theorem 362 

9.1 Introduction 363 / 9,2 Area Ideas 364 / 9.4 Area 
Postulates 368 / 9.4 Area Formulas 372 / 9.5 
Pythagorean Theorem 379 / Chapter Summary 385 / 
Review Exercises 386 

Chapter 10 

Similarity 390 

10.1 Introduction 391 / 10.2 Proportionality 392 / 10.3 
Properties of Proportionalities 396 / J 0.4 Similarities 
Between Polygons 401 / 10.5 Some I-ength Proportionalities 
409 / 10,6 Triangle Similarity Theorems 415 / 10.7 
Similarities in Right Triangles 422 / 10.8 Some Right Triangle 
Theorems 429 / Chapter Summary 436 / Review 
Exercises 437 

Chapter U 
Coordinates in a Plane 440 

11.1 Introduction 441 / 1 1.2 A Coordinate System in a 
Plane 442 / 1 L3 Graphs in a Plane 447 / 11.4 Distance 
Formulas 452 / 11.5 The Midpoint Formula 458 / 11.6 
Parametric linear Equations 464 / 11.7 Slope 473 / 1 1.8 
Other Equations of Lines 481 / 11.9 Proofs Using Coordinates 
494 / Chapter Summary 506 / Review Exercises 507 



xii Contents 



Chapter 12 

Coordinates in Space 510 

12.1 A Coordinate System in Space 511 / 12.2 A Distance 
Formula 519 / 12.3 Parametric Equations for a line in 
Space 523 / 12.4 Equations of Planes 526 / 12.5 
Symmetric Equations for a Line 539 / Chapter Summary 
544 / Review Exercises 544 

Chapter 13 

Circles and Spheres 546 

13.1 Introduction 547 / 13.2 Circles and Spheres: Basic 
Definitions 547 / 13.3 Tangent Lines 544 / 13.4 Tangent 
Pknes 567 / 13.5 Circular Arcs, Arc Measure 576 / 13.6 
Intercepted Arcs, Inscribed Angles, Angle Measure 
584 / 13.7 Segment*: of Chords, Tangents, and Secants 
596 / Chapter Summary 607 / Review Exercises 608 

Chapter 14 

Circumferences and Areas of Circles 612 

14.1 Introduction 613 / 14.2 Polygons 614 / 14.3 
Regular Polygons and Circles 622 / 14,4 The Circumference 
of a Circle 634 / 14.5 Areas of Circles; Arc Length; Sector 
of a Circle 643 / Chapter Summary 636 / Review 
Exercises 658 

Chapter 15 

Areas and Volumes of Solids 662 

15 .1 Introduction 663 / 15.2 Prisms 664 / 15.3 
Pyramids 672 / 15.4 Areas and Volumes of Prisms and 
Cylinders 680 / 15.5 Volumes of Pyramids and Cones 
688 / 15.6 Surface Areas of Spheres and Volumes of 
Spherical Regions 697 / Chapter Summary 702 / Review 
Exercises 704 



Contents xiii 



Appendix A-l 

list of Symbols A- 1 

List of Postulates A-2 

List of Definitions A-5 

List of Theorems A-20 

Table of Squares, Cubes, Square Roots, Cube Roots A-42 

Glossary G-] 
Index 1-1 





Chapter 




Sam Folk /Mankmeycr 



Points, 
Lines, and 
Planes 



1.1 INFORMAL GEOMETRY 

Geometry began informally more than 2(KX) years ago in Babylon 
and Egypt. Tlie meaning of geometry is, literally, "earth measure- 
ment"; hence it is not surprising thai ruilv knowledge of the subject 
was concerned largely with measurements of lengths, areas, and vol- 
umes. Such knowledge was a necessity, especially to the aneient Egyp» 
tians who almost annually were forced to restore to their river-side 
farms the boundary markers which were washed away by heavy flood- 
ing of the River Nile. 

Then, as well as now, men learned by experience. The ancient 
Egyptians developed many rules-of-thmnb for surveying fields and 
roads and for making calculations related to building dwellings and 
pyramids. These rules were based on many observations, intuition, and 
reasoning. As long ago as 500 B*G men knew how- to find the areas of 
rectangles, triangles, and trapezoids, but they had made little progress 
in formal geometry, that is, in developing geometry as a system that 
explains why the rules worked. 



Points, Lines, and Piano* 



Chapter l 



As a school subject today geometry is bolh informal and formal 
In the elementary schools geometry is largely informal. It is physical 
geometry. Students work with physical objects or with pictures thai 
represent physical objects, General statement 1 ? or rules are based on 
intuitive reasoning and inductive reasoning. Intuitive reasoning is what 
might be called common sense, or, as some might say, reasoning in a 
hurry. Inductive reasoning is reasoning based on numerous examples. 
Wc discuss an example of each. 



Intuitive Reasoning 

Let us take for granted that we know what is meant by a triangle. 
Figure 1-! shows a right triangle A ABC with right angle at C 




figure M 

Suppose that the lengths of the sides AO and BC are b inches and 
a inches, respectively. What b the area of the triangle? We say "area 
of the triangje/' but we really mean the area of the figure or region 
that consists of the triangle and its interior. In earlier mathematics 
classes we have learned that the area is \ab square inches. Suppose 
someone asks why. One answer might be: *"It is intuitively obvious. 
You can see it by looking at a picture of a rectangle with sides of lengths 
a and h and with one diagonal drawn." (See Figure 1-2.) If the person 
who asked why responds with "Now I see why," he is responding to 
an intuitive feeling of the "tightness" of tilings rather than to a logical 
argument or to inductive reasoning. This is an example of intuitive 
reasoning. 




Figure 1-2 



1.1 Informal Geometry 



Inductive Reasoning 

Ixl us take for granted that we know what is meant by an angle 
and its measure. We record the measures of angles as numbers, omit- 
ting the degree symbol. This is consistent with our formal point of view 
regarding measure developed later. Suppose that each student in an 
elementary geometry class is given a set of plastic triangles numl Hired 
1, 2, 3, 4, as suggested by Figure 1-3, and a protractor. The instructions 







direct the students to find and record the measures of the angles of the 
triangles and then to Hnd the sum of the measures for each triangle as 
in the following table. The object of the lesson is to make it plausible 
that the sum of the measures of the angles of a triangle is 180, If the 
students reason that this is a correct conclusion on the basis of the 
measurements they have made, this is an example of inductive 
reasoning. 



mLA 



mlB 



m t C 



Triangle 1 


90 


32 


58 


180 


2 


75 


26 


78 


179 


3 


95 


25 


62 


182 


4 


65 


25 


90 


180 



Points, Lines, and Planes : i ,ip-.er 

1.2 THE IDEA OF FORMAL GEOMETRY 

Suppose that as a result of inductive reasoning we htive made a 
general statement about all triangles, or about all rectangles, or about 
all circles. We think that the statement is true, but how can we know 
for certain? Is it possible to know something for certain about all tri- 
angles? It would seem necessary to test every triangle, which is clearlv 
impossible. Therefore a different approach is required, the formal 
approach. 

The formal approach involves deductive reasoning, that is, logical 
arguments by which general statements are obtained from previously 
accepted statements- Two examples follow. 

Deductive Reasoning 

Example Suppose that you are given a set of plastic convex quadri- 
laterals numbered 1, 2, 3 t . . . , 10 as suggested by Figure i-4 Suppose 
that there are really ten of them, although only four of them are shown. 




Figure m 

Suppose also that you know that the sum of the measures of t lie angles 
of every triangle Is 1 80. Perhaps you know this through the example 
of inductive reasoning in Section LL At any rate you know or are told 
this. This knowledge about all triangles is considered as "given" in the 
situation of this example. The instructions are to find the sum of the 
measures of the angles of each quadrilateral without making any meas- 
urements using a protractor. 

A quadrilateral is convex only if it has the property that if either 
pair of its opposite vertices is joined by a segment, called a diagonal 
of the quadrilateral, then all of that diagonal except its endpoints lies 
inside the quadrilateral. Figure I -5a shows a convex quadrilateral. In 
Figure 1 -5b the segment PH, except for the points P and R, lies outside 
the quadrilateral. Hence PQRS is not a convex quadrilateral. 

A picture of a quadrilateral and one of its diagonals provides a clue 
for finding the sum of the measures of the angles, (See Figure 1-6.) 
Quadrilateral ABCD may be considered representative of any convex 
quadrilateral. From this flgtirc we see that the sum of the measures of 



1,2 The Idea of Formal Geometry 5 




(a) Convex 



(b) Not convex 



Figure 1-5 




Figure 1-6 



the angles of a quadrilateral is the sum of the measures of the angles 
of two triangles. Therefore the sum of the measures of the angles of a 
quadrilateral is 2 * 180 or 360- Without using a protractor we decide 
that the sum of the measures of the angles of each quadrilateral is 360. 
We arrive at this conclusion by deductive reasoning. 

Let us examine this example carefully so that we may understand 
more clearly the nature of deductive reasoning. We took some things 
for granted or as given. From these we deduced the answer, or con- 
clusion. The given statements are called hypotheses, or collectively, 
the hypothesis. Thus the essence of deductive reasoning is to obtain a 
conclusion from a hypothesis- The hypothesis may or may not be true. 
If it is true in a given situation and if the reasoning involved in reaching 
die conclusion is correct, then the conclusion is true in that situation. 

Let us look at this idea expressed in symbols. Suppose that wc are 
given the hypothesis, denoted by If, and we want to establish the con- 
clusion, denoted by C* What we want to prove is not "C" but "If ff, 
then C." Many statements in mathematics are in this "If H t then C" 
form- How can such a statement be useful? It is useful in any situation 
where the "H " is true. Because if we know "H" is true and if we know 
"If IT, then C " is true, we also know that "C" is true. In the problem 
about quadrilaterals, the hypothesis and the conclusion may be identi- 
fied as follows; 

H: ABCD is a convex quadrilateral. 

C: The sum of the measures of the angles of ABCD is 360. 



6 Prints, Lines, and Planes 

If ABCD is a quadrilateral like the one in Figure T -5a, then '*H " is 
true, and since "If H, then C" is true, it is correct to conclude that "C" 
is true. If ABCD is a quadrilateral like the one in Figure l-5b, then 
"H" is false, and although "If ti % then C " is true, it is incorrect to con- 
clude that "C " is true. 

Example This example in which deductive reasoning is used to es- 
tablish a general statement concerns a property of the natural numbers, 
1, 2, 3, . . . . Undoubtedly, you know an even number is a number that 
is 2 times an integer. Thus x is an even number if there is an integer ij 
such that x = 2y. Some even numbers are 2, 16, 168, and 2466. If you 
add any two of these numbers, you get an even number. Is it then true 
that the sum of any two even numbers is an even number? Of course 
it is. Let us prove it, however, by using deductive reasoning. 

H: x and y are even numbers. 
C: x + y is an even number. 

Our task is to prove: If H, then C. 

Proof: Since x is even, there is an integer u such that x = 2n. Since 
y is even, there is an integer c such that y s 2c. Then 

x + \j = 2u + 2c 

and, using the distributive property, we get 

x + y = 2{u + v). 

But u -+■ 15 is an integer. Therefore, since x -\-yis2 times an integer, 

x + y is an even number. 

Notice that we did not prove that x is even or that y is even. We 
proved that if x and y are oven numbers, then *-f y is an even number. 

As we said, formal geometry involves deductive reasoning. An im- 
portant feature of a formal geometry is its structure or arrangement. 
Hie geometry of this book is elementary Euclidean geometry, carefully 
arranged so that we can see how the various parts fit together and how 
some things depend on other things. Formal geometry might be 
thought of as geometry for the lazy person. In formal geometry we 
prefer a general statement that tells something significant about all 
triangles rather than a hundred statements about a hundred triangles. 

Our starting point in formal geometry is a set of statements about 
some of the simplest objects of geometry. We do not try to tell what 
these objects are by definitions. Definitions would involve other words 
that we would need to define in turn, and so we accept some concepts 



1,2 The Idea of Formal Geometry 

as basic and undefined. We might decide, for example, to start with 
triangles because almost everyone lias an idea of what a triangle is. But 
do we actually know what a triangle is? A triangle is made up of three 
segments. The notion of segment is more basic than the notion of tri- 
angle. Every segment is a set of points. The idea of a point is more 
fundamental than the idea of a segment. Every segment is a part of 
some line. Perhaps a line is simpler to think about than a segment 

In formal geometry we usually consider points, lines, and planes 
as the basic building blocks. We do not define these words. How then 
can we be sure that wc know anytlung at all about them? On the basis 
of our experience with physical objects we identify the most basic 
properties that points, lines, and planes have in relation to one an- 
other. We formulate these as statements that we accept without proof. 
We call these statements postulates. 

The foundation for formal geometry, then, is a set of statements, 
the postulates, which we accept without proof. "The postulates are 
statements about the basic objects in geometry. We agree that what 
we know about these objects is what the postulates say, and nothing 
more, at least at the start. What else can we possibly know about these 
objects? We can know what %ve formally assume in the definitions and 
what we deduce by logical reasoning. 

As an example suppose that we start with the following six postu- 
lates and one definition. (This is not for keeps— just for this example!) 
Our notations arc self-explanatory. 

POST U L ATE A Every line i g a set of p oints and contains at least 
two distinct points. 

POSTULATE B Every segment is a subset of a line and contains 
(besides other points perhaps) exactly two distinct points called its 
endpoints. 

POSTULATE C If A and B are any two distinct points, there is 

exactly one line AB containing A and B, and exactly one segment AB 
with A and B as endpoints. 

POSTULATE D Every plane is a set of points and contains at 
least three points that do not all he on one line, that is, three noncol- 
linear points. 

POSTULATE E If A, B, C arc three noncollinear points, then 
there is exactly one plane containing A, B, and C. 

POSTULATE F If a plane contains two distinct points A and B, 
then it contains the segment AB. 



S Points, Unes, and Planes 



Chapter 1 



Definition If A t B t C arc any three noncolHnear points, then 
the onion of the three segments AB, BE, and ^A is a triangle; 
we denote it by A ABC. 



In this example, what do we know for sure? We know what is said 
in these seven statements, the six postulates and the definition. What 
else do we know? We know anything else that we deduce by logical 
reasoning from them. We shall deduce one statement and call it a 
theorem. 

THEOREM Every plane contains at least one triangle. 

Proof; Let a (read "alpha") be any plane. Then a contains three non- 
colltiiear points (Postulate D)j call them A, fl, C. Then there is exactly 
one segment AB with endpoints A and B t exactly one segment BC with 
endpoints B and Q and exactly one segment Cli with endpoints C and 
A (Postulate C). Then there is a triangle AABC (by our definition). 
Each of the segments AB, B£, CA lies m a {Postulate F). Therefore 
AABC, which is their union, lies in a and the proof is complete. 

In proving a theorem it is usually a good idea to include one or more 
inures l.i> suggest the given .situation. In this Mliialirn we start witli 8 
plane a as suggested in Figure 1-7, Next we reason deductively to get 
duee noncollincar points A, B, C in a as suggested in Figure 1-8, 
Finally, we reason deductively to show that there is a triangle AABC 
and that it lies in a as suggested in Figure 1-9* 




Figure 1-7 



Our theorem may seem trivial to you. Our goal, however, was sim- 
ply to show an easy example of a theorem obtained by deductive rea- 
soning from a set of postulates. 



1,2 The Idea of Formal Geomttry 



EXERCISES 1.2 

In Exercises 1-0, a situation is given and a question is asked Obtain an an- 
swer in each case using intuitive reasoning, 

1. Given ihc two numbers 

x a 21.3 + 27.4 and y = 28.5 + 16.2 + 5.9, 

which is larger, x or y? 

2. Given the two numbers 

p = (1.6X754) and q = (L6X89G), 

is pq greater than, equal to, or less than (2)(76SX896)? 

3. The following figure shows a square and a parallelogram with side 
lengths unci angle measures as labeled. Do these figures have equal 
areas? If not, which one has the larger area? 




4 The figure below shows two parallel lines. Is there a plane containing 
these two lines? 



«- 



5. The figure shows a rectangular box with the vertices labeled. Is there a 
plane containing the four points ft, G, C* A? 




6. Given the situation of Exercise 5, is there a plane containing the four 
points A, R, C t H? 



7. 


H: 




C; 


8. H: 




C: 


9, 


H: 




C: 


10. 


H-. 




C: 


TL 


H: 
n. 


12. 


H. 




C: 



10 Points, Lines, and Planes Chapter 1 

■ In Exercises 7-12, a hypothesis H and a conclusion C are given. State 
whether you think the conclusion follows logically from the hypothesis. Be 
prepared to defend your answer. 

x is an odd number. 
x 2 is mi odd number. 

ar is a multiple of 3. 
a -2 is a multiple of 6. 
X is a multiple of 3, 
x 2 Is a multiple of 9. 

A ABC is a right triangle with sides of lengths 3 in. 3 4 in. ; and 5 in. 

The area of &ABC is 6 sq- in. 

p and q are two distinct intersecting planes. 

The intersection of p and <] is a line. 

S is a sphere and p is a plane that intersects 55. 

The intersection of S and p is a circle or a point. 



13 THE IDEAS OF POINT, LINE, AND PLANE 

Every day in the world around us we observe objects of different 
sizes and shapes. We notice that some of these objects have corners, 
edges, and sides, and that some of their parts are "straight," some are 
"flat," and some are "round." Touching and seeing certain objects help 
ns to classify them according to their characteristics. 

Tn arithmetic the idea of a number is a mathematical idea that grew 
out of a need to classify certain sets according to how many objects 
they contained. But. no one has ever seen or touched a number. In ge- 
ometry the ideas of point, line, and plane arc mathematical ideas that 
grew out of a need to classify certain sets of figures and to measure thei r 
boundaries or the regions bounded by the figures. But no one has ever 
seen or touched a point, a line, or a plane. Just as in arithmetic you 
studied numbers and the operations on them, in geometry you will 
study points, lines, and planes and how they relate to one another. 

In the same way that deductive reasoning must be based on certain 
assumptions (postulates) that we accept without proof so must our 
definitions be based on certain terms that we make no attempt to de- 
fine. Tliis is necessary in order to avoid "circular" definitions, that is, 
a chain of definitions which eventually comes back to the first word 
being defined. If a person does not know the meaning of any of the 
words in the chain, the definition is of no value to him. For example, 
in defining the word "dimension" one dictionary uses the word "mag- 
nitude." It defines "magnitude" In terms of "size." When looking up 



1.3 Th» Idtai of Point, Line, and Plane 1 1 

"size" in this same dictionary, it gives "dimension or magnitude," 
Thus, if someone does not know the meaning of any of the terms mag- 
nitude, size, or dimension, the dictionary is not very useful. Somewhere 
in the cycle it is necessary to know the meaning of a word based on 
experience. 

The basic undefined terms in our geometry are p>trrf T line, and 
plane. Our postulates give these terms the meaning that we wish them 
to have. You undoubtedly already have an intuitive feeling for the con- 
cept of a point, a line, or a plane. We can think in a vague fashion of a 
point as having "position" but no "size"; of a line as being "straight;" 
having "direction," but no "width"; and a plane as being "Hat" but 
having no "thickness." When we "mark a point" or "draw a line" on 
our paper or on the chalkboard, we are merely drawing a picture of 
what we think a point or a line should be, These pictures or figures help 
us to see and discover some of the relationships that exist among points, 
lines, and planes and help us to keep these relationships straight in our 
minds. However, any deductions that we reach must be justified 
strictly on the basis of our postulates, definitions, and theorems and not 
on what appears to be true from a figure. As we progress in our study 
of geometry, we will use figures more and more freely. If our figures 
are drawn carefully enough, they generally « ill not give us false infor- 
mation or lead us to false conclusions. 

Indeed, trying to deduce al the theorems and work all the prob- 
lems in this text without drawing any figures would be a tedious and 
difficult task. It would be somewhat like a carpenter attempting to 
construct a house from memory, without the aid of any blueprints or 
floor plans. The resulting structure might be quite different from the 
house he planned to build. 

We use figures frequently to help explain what our postulates, defi- 
nitions, and theorems say. You are cneouruged to do the same. When 
it is practical, you should restate a theorem or a problem in terms of a 
figure that shows the relationships that arc given in the theorem. If you 
are careful not to include in the figure any special properties that are 
not given in the theorem, then the figure should be a valuable aid in 
understanding the theorem and also in proving it. 

EXKRCISES 1.3 

1. Name three physical objects that convey the idea of a point. 

2. Name three physical objects that convey the idea of a line. 

3. Name three physical objects that convey the idea of a plane. 



12 Points, Lines, and Planes 



Chapter 1 



Exercises 4-9 refer lo the cube shown in Figure 1-10. Answer each question 
in Exercises 4-7 with the word point, lino, plane, or space, 

4. Each cornei' (vertex) of the cube suggests 
ft!®. 

5. Each side (face) of the cube suggests a [7|. 

6. Each edge of the cube (such as ATS) suggests 

7. The interior of the cube is a subset of [Tj. 
8* How many vertices Joes the cube have? How 

many faces? How many edges? 
&. Let V represent the number of vertices, E 
the number of edges, and F the number of 
laces. Compute V — E + F for the cube. 




Figure MO 



In Exercises 10-17, copy and complete the table on the following page for 
the pyramids and prisms shown below. 






14 





15. 




1,3 The Ideas of Point, Line, and Plane 13 




17, 



i«ZI> 




E 



10. 



E + F 



12. 

13. 

14, 
15. 
10. 
17. 



18. I5id you get the same number for V — £ + Fin Exercises 9-17? If you 
have counted correctly, V — E ■+■ F equals 2 in each case. Figures like 
those in Exercises 1CM7 are called polyhedrons. On the basis of your an- 
swers for Exercises 9-17, do you think that V — E + F = 2 for every 
polyhedron? Compute V T jE, F, and V — E + Ffor the star-shaped poly- 
hedron shown in Figure 1-11, (You can construct this polyhedron by 
gluing together 16 triangles and 2 squares.) 




Figure MI 



14 Points, Lines, and Planes Chapter 1 

1.4 SETS 

In our development of formal geometry we shall often speak of 
sets of points, or sets of lines, or sets of planes, as well as sets of num- 
bers. Indeed, we shall agree later in this chapter that every line is a set 
of points and that every plane is a set of points. Although you are fa- 
miliar with the language of sets, we s h all review it briefly. 

Recall that a set is simply a collection of objects. The objects may 
be numbers, people, points, or whatever. The objects of u set are called 
the members or the elements of a set. The symbol € indicates that 
an object is an clement of a given set. Thus, to indicate that the num- 
ber 2 is an clcmenL of the set {1, 2, 3, 4), we write 

2 €{1,2, 3, 4}. 

Wc read the expression 2 £ {1, 2, 3, 4} as "2 is an element of the set 
{ 1, 2, 3 t 4}" or simply as "2 is in { 1, 2, 3, 4}/' Similarly, if point A is an 
element of the set of points in line J, we write 

A£l 

and read this as "'point A Is on line h" 

It is often convenient to use set-builder notation to indicate the 
members of a particular set For example, 

{x : x is an integer and —2 < x < 4} 

is just another way of naming the set { — 1, 0, 1, 2, 3}, We read the ex- 
pression {x : x is an integer and —2 < x < 4} as "the set of all num- 
bers x such that X is an integer and X k greater than —2 and less than 
4." If we wished to include —2 and 4 in this set, we would write 
— 2 < x < 4 rather than —2 < x < 4. 

Example 1 Let R represent the set of all real numbers. Suppose thai 

wc wish to picture (graph) the set 

C = {x : x € H and -3 < x < 3} 

on a number line. Another way of writing the expression —3 < x <[ 3 
is x ^> — 3 and x < 3. 

The number line in Figure 1-12 pictures the set 

A = (s : x € R and x > -3}. 



■o— ! i 



-6 -5 -4 -3-2-1 1 2 3 4 
Figure 1-12 



1.4 Sets 15 



Note that the point corresponding to the number —3 has been circled 
to indicate that —3 is not an element of this set. 
The number line in Figure 1-13 pictures the set 

B = {x : x £ R and x ^ 3}. 

Why is the point corresponding to the number 3 circled and shaded in 
the figure? 

B 



«- 



■S -5-4-3-2-1 I 2 3 4 5 fi 

Figure 1-1*3 



It is clear from the two number lines in Figures 1-12 and 1-13 that 
the numbers that are common to both sets (those that belong to both 
A and R) are the real numbers between —3 and 3 and including 3. 
Tills is shown on a single number line in Figure 1-14. 

c 

4 o ♦ — ! ► 

-6-5-4-3-2-1 J 2 3 4 5 6 

Figure 1-14 

We call the set shown in Figure 1-14 the intersection of the two 
sets in Figure 1-13. The intersection symbol D is used in forming a 
symbol to denote the intersection of two sets. Thus to indicate the in- 
tersection of the two sets 

A = {x : x £ H and x > -3} 
and 

B = {x : x £ R and x £ 3} 

we write A n B and we see that tills intersection is the set 

C = {x : x £ R and -3 < x <, 3}. 

rherefore C = A P B. 

In set language the connective "and" is related to intersection. 
"Thus the elements that belong to sets A and B in Example 1 are those 
elements that are common to the two sets; in other words, the inter- 
section of the two sets is the set C. 

It may happen that the intersection of two sets is empty, meaimi Li 
that the two sets have no elements in common. The symbol indicates 
the empty or null set For example, since the set E of even integers and 
the set O of odd integers have no integers in common we write 

En 0=0. 

In formal geometry, when we say that one set intersects another 
set, we mean that the two sets have at least one element ra common. 



16 Points, Lines, and Planes Chapter 1 

In this case, the intersection of the two sets cannot be the empty set. 
For example, consider the three sets 

D = {2, 3. 5, 7}, 
F = {0, 4, 6, 9}, 
G = (3, 5, 8, 10}. 

Sets D and F do not intersect since they have no elements in common, 
and we write DOF= , However, sets D and G do intersect and 
we write DrC=(3,5). 

Tn listing the elements of a set, as for D, F, and G, the order in which 
the elements appear is not important. Thus, if U = {2, 3, 5, 7} } then 
we may also write D = (2, 5, 7, 3}, The intersection of two sets is in- 
dependent of order, too. Thus, if A and B are any two sets, then 

a n n = b n a. 

Example 2 Let set = {x : x € J* and x < -3 or * > 3}, Let us 
picture the sets 

A = {*:x€Rand*< -3} 
and 

i?={x:x€fiandx>3} 

separately on two parallel number lines. (Note: A and B are not the 
same sets as in Example 1.) 

The number line (a) in Figure 1-15 shows the graph of the set 

A = {x : *€ Rand*< -3} 

and the numt>er line (b) shows the graph of the set 

8= {x :x€Randx>3}. 

A 

(a) 4 



0>) 



















_| 










i 




-6 


-ft 


-* 


-3 


-2 


-1 





i 


2 


3 


4 a 


u 




4- 






















B 




-► 



-6 -5 -4 -S -2 -t 
Figure 1-15 



Those numbers that belong to set A or set B are the numbers that he 
to the left of -3 or to the right of 3. Figure 1-16 on page 17 shows this 
on a single number line. 



1.4 SMs 17 

D 
4 i . > 

-6 -5-4-3-2-1 1 a U i 5 6 

Figure I- 10 

We call the set shown in Figure 1-16 llic union of the two sets in 
Figiire 1-15. The union symbol U is used in forming a symbol to denote 
the union of two sets. Thus to indicate 'the union of two sets 

A = {x : x £ R and x < -3} 
and 

B = {% : % € R mix > 3} f 

we write A U B and see that this union is the set 

D= {x : *EHand*< -3ori>3). 

Therefore D = A U B. 

Example 2 illustrates how the connective "or" is related to unions 
in the language of sets. Thus the elements that belong to sets A or B 
are those elements that are cither in set A or in set B or in both. The 
set of all elements that belong to A Of B is the union of the two sets 
A and B which, we have seen, is the set D. 

The union of two sets is independent of their order. Hence, if 
A and B are any sets, then A J B = B J A, 

The ideas of union and intersection are extended easily to apply 
to any number of sets. For example, if A, B, and C are sets, then 
A n B n C is the set of all elements each of which is in aU three of 
the sets A. B, and C Describe the intersection of the two sets A and B 
of Example 2L 

Example 3 Let A = { l t 2, 3, 4, 5} and B = (3, 4, 5, 6, 7}. The union 
of A and B is the set of numbers in A or B, and we write 

AU B= {.1, 2, 3* 4, 5, 6, 7}. 

It should be noted that if a number is in set A, or in set B, or in both 
set A and set B, then this number is in A J B. Thus 1 and 2 belong 
in A U B because 1 and 2 are in set A. The numbers 6 and 7 belong 
in A U B because 6 and 7 are in set B. Finally, the numbers 3, 4, and 5 
belong in A J B because 3 T 4, and 5 are in both set A and set B, The 
intersection of A and B is the set whose elements are common to both 
sels A and B, and we write 

AH B= {3,4.5}. 



18 Points, Lines, and Planes Chapter 1 

Recall that a set A is a subset of set B if every element of A is also 
an element of B. (Write A C B for "A is a subset of B") Thus, if 

A = {«, b), then each of the sets («}, [h], {a, b} t and is a Subset 
of A. 



EXERCISES 1.4 

L Let A = {x : x£ flandx< lJandB = {x : x£ R andx > -I). 
(a) Graph the set A. the set B, and the set A (1 B on separate number 

lines, 
(h) Use set-builder notation to indicate the intersection of sets A and H. 
(c) Use set-builder notation to indicate the union of sets A and B. 

Kxercises 2-8 refer to the sets 

A= {-3, -1, 1,3},B= (-2.. 0,2, 4}, and 
C={-2, -1,0,1,2}. 

2* Find A n B. A Find A U CL 

3. Find A n C. 7. Find A UBUC 

4. Find BnC. 8. Find (AtBJfl C. 

5. Find A U B, 

9. The figure below shows the graph of a set of numbers on a real number 

hue. Use set-builder notation to describe this set. 



-7 -6 -5 -4 -3 -2 -1 



> 



10, The figure below shows the graph of a set of numbers on a real number 
line. Use set-builder notation to describe this set 



-"- 



-6 -5 -4 -3 -a -1 



1.5 CONJUNCTIONS AND DISJUNCTIONS 

Similar to the relationships of two sets and their union and inter- 
section are the relationships of two statements and their conjunction 
and disjunction. In mathematics it is very important to understand 
clearly the relationship of truth in given statements to truth in state- 
ments formed from the given ones. In this section we consider the rela- 
tion of two statements to the statements formed from them by con- 
necting them first by "and" and then by "or." 



1.5 Conjunctions and Disjunctions 19 

Suppose that p is a statement and that q is a statement. (By a state* 
ment we mean a sentence that is cither true or false.) We call the state- 
ment "p and (/"the conjunction of the statement p and the statement 
q> For example, if p is the statement 

4 = 4 
and q is the statement 

4 = 5 t 

the conjunction of p and q is the statement 

4 = 4 and 4 = 5. 

We agree to call a conjunction of two statements true if and only if the 
statements are both true. 

► If p is true and q is true, then p and q is true; if p and q is tru e, 
then p is true and q is true. 

Thus the conjunction 4 = 4 and 4 = 5 is false because the state- 
ment 4=5 is false. On the other hand, the conjunction 4=1 unci 
5 = 5 is true. Of course, if both of the statements p and q are false, 
then the conjunction p and q is false. 

Let us see how the conjunction of two statements is related to the 
intersection of two sets. In Example 1 of Section 1.4, we considered 
the intersection of two sets 

A = {« i x £ R and x > -3} and B = {x : x £ R and x < 3}. 

We saw that the intersection of these two sets is the set of aE numbers 
that belong to both A and B. In other words, if X is a number, then it is 
true that x is an element of A and of 8 (x £ A D B) if and only if it is 
true that x is an element of A and it is true that x is an element of B. 
For example, 2 £ A and 2 £ B; hence 2 £ A P B. On the other hand, 
5 £ A and 5 £ B (5 is not an element of B), hence 5 £ A n J3. 

Thus if the statements i£A and x £ B are true, then the conjunc- 
tion of these statements (x £ A n B) is also true. But if either of the 
statements x £A and * £ B is false, then * € A n J? is false. 

Now consider the statement p or q where p and q are statements. 
We call the statement p or q the disjunction of the two statements p 
and C/. For example, if £? is the statement 

4 = 4 
and f/ is the statement 

4 = 5, 

the disjunction of p and q is the statement 4 = 4 or 4 = 5. 



20 



Points. Lines, and Planes 



Chapter I 



We agree to call the disjunction of two statements true if and only if 
either, or both, of Ihc two statements is true, 

► If /> is true and q is true, then p or q is true; if p is true and q 
is false, then povq is true; if /> is false and q is true, then p or 
q is true; if p or q is true, then either p is true and q is false, or 
p is false and q is true, or p and q are both true. 

Thus the disjunction 4 = 4 or 4 = 5 is considered to be true be- 
cause 4 = 4 is true. Similarly, the disfunction 4 as 4 or 5 s 5 is t me 
since the statements 4 = 4 and 5 = 5 are both true. On the other 
hand, the disjunction 4 = 5 or 7 < 5 is False since I Kith of the state- 
ments 4 = 5 and 7 < 5 are false. 

Let us see how the disjunction of two statements is related to the 
union of two sets. In Example 2 of Section 1.4 we considered the union 
of the two sets 

A = (x : x € B and x < -3} and B = {x : x £ R and * > 3}. 

We saw that the union of these two sets is the set of numbers belong- 
ing to either A or B. In other wortfc. if x is a number, then it is true that 
i is an element of A or B (x € A U B) if and only if at least one of the 
statements x is an element of A and x is an element of B is true. For 
example, 5 € A, but 5 € B; hence 5 C A U R On the other hand, 
2 g A and 2 £ B; hence 2 t A U B, 

Thus, if at least one of the statements x £ A and x £ B is true, then 
the disjunction of these two statements (x £ A U B) is also true. But 
if both of the statements x € A and x £ B are false, then the statement 
x 6 A U B is false. 

FAERCISES 1.5 

Exercises 1-6 refer to Figure J -17. Imagine that each of the four lines a, b, 
c.din the figure is a set of points. 



FJeuru I- 17 




3 .5 Conjunctions and Disjunctions 21 

L The set {£} is the intersection of which two lines? 
2. What is the intersection of h and c? 
3* What is the Intersection of the three sets a t h, and rf? 
4 Does « fl fc = a P fi? 

5, What is a H c? 

6. What is the intersection of the four sets a, b, c, and d? 

The following table shows the possible truth values (T for true, F for false) 
that can be assigned to the statements p and q. In Exercises 7-10, copy and 
complete the table for the statements p and q and p or q. 



p and q p or q 



7. 


T 


T 


T 


8. 


T 


F 


F 


ft. 


F 


T 


CD 


10. 


F 


F 


DQ 



11. If x £ A and x £ B t is ran element of A H B? Of A U B? 

In Exercises 12-15. P and Q represent sets. Copy and complete the table 
using T for true or F for false. 





*€ P 


x€Q 


x 6 P n Q 


XC p \J Q 


12. 


T 


T 


T 


7] 


13. 


T 


F 


a 


m 


14. 


F 


T 


3 


3 


15. 


F 


F 


13 


F 



16. Two sets arc said Lo be equal if they contain exactly the same elements. 
If A n {x ; ar is an integer and - 1 < % < 5} and B - {0, 1, 2 S 3, 4), 
does A = B? Why or why not? If C = {2. 3, 0, 1 , 4 } } does R = C? 

17, In Exercise 1.6 is A a subset of B? Is B u subset of A? Is 5 a subset of C? 
Is C a subset of B? 

IS. If A C B and B C A are both true, what can you conclude? 

19. If A and B arc sets and A = B i does B = A? 

20. If A and B arc sets and A C B, is B C A? 



22 Points. Lines, and Planet Chapter 1 

21. (a) List all the subsets of the set { 1, 2, 3}. (Remember that the empty 

set is a subset of every set.) 
(b) How many distinct subsets does a set co nsisting of three elements 
have? Four elements;? n elements? 

22. If A = and & = {3, 4, 5, 6}, what is A H B? A J F? 

23. (a) Given that x is a number such that x 2 = 16, find one possible value 

for x. Is there another possible value for x? What is the solution set 
of the equation x 2 = 16? 
(b) Write die solution set of the equation x 2 + 7 = 56. 

■ In Exercises 24-29, write the solution set of the equation. 

24. 3a - 5 = 13 27. 2(* + 3) = 5a - 8 

25. 5(2* - 9) = 10 28. x* - 5 = 31 

26. 3(2* - 7) = -t + 9 29. x 2 - 2x - 15 = 

30. If F = f (x, y) : x £ H, tj C R. and y = 2x - 3}, we find that when 
i=2 T y = 2'2 — 3 = 1. Thus the ordered pair (2, 1} is an element 
of the set F. To check this, note that "2 £ R, 1 € R, and 1 = 2 ■ 2 - 3" 
is a true statement. This is the statement you get if you replace x by 2 
and y by 1 m the sentence which follows the colon in the set-builder 
symbol 

(a) What value of tj is paired with x when x = ^? 

(b) Find five more ordered pairs of numbers that belong to the set F. 

31. Describe the solution set of the equation 2(x — 2) = 2x — 4. 

32. Describe the solution set of the equation 2{x — 2) = 2x - 5. 

1.6 THE INCIDENCE RELATIONSHIPS 
OF POINTS AND LINES 

We are now ready to begin the formal development of geometry. 
We do not state all the postulates of our formal geometry at once, but 
rather introduce them throughout the text as needed. In this chapter. 
we state the first eight of que postulates. These postulates are con- 
cerned with points on lines, in planes, and in space; lines through 
points, in planes, in space ; and planes through points, through lines, 
and in space. Figure 1-18 shows a picture of a line I and a point F. As 
the figure suggests, point F is on line / and line / passes through point F. 
Since t is a set of points and P is an element of this set, it is correct to say 
that F is "in" I or that F is a point "of" L However, we usually say that 



Figure 1- 



1,6 Relationships of Points and Lirv« 23 

Pis "on" I and that 1 "passes through" P, Another way of saying this is 
to say that the point and line are incident, it is for this reason that our 
first postulates, which we state in this section and in Section 1.7, are 
called Incidence Postulates: 

For our first postulates we draw on our experience with physical 
representations of points and lines. When you use your ruler or 
straightedge to draw a line through two points, such as P and Q in Fig- 
ure 1-19, you are, of course, actually working with physical representa- 
tions of points and lines and not the abstract points and lines of our 
geometry. 




Figure 1-19 

We decide what we are going to accept without proof about geo- 
metric points and lines by looking at the real world. In fact, we need 
only look at Figure 1-19 to arrive at our first three postulates. First, 
however, wc state two definitions. 



Definition 1.1 Space is the set of all points 

Definition 1.2 The points of a set are collinear if and only 
if there is a line which contains all of them. The points of a 
set are n uncoil in ear if and only if there is no line which con- 
tains all of them. 



Plane Postulates of Incidence 

POSTULATE 1 (The Tiiree-Point Postulate) Space contains 
at least throe noncollinear points. 

POSTULATE 2 (The Line-Point Postulate) Every line is a set 
of points and contains at least two distinct points. 

POSTULATE 3 (The Point-Line Postulate) For every two 
distinct points, there is one and only one line mat contains both points. 



24 Points. Lin«. and Planes Chapter 1 

Postulate 3 is sometimes shortened to read "two points determine 
a line." When we use the word "determine" here, we use it in the 
sense of Postulate 3, which states that, given two distinct points, there 
is exactly one line which contains them. Similarly, when we say that 
"three noncolliiicar points determine three lines," we mean that there 
are exacdy three lines each containing two of the three points. 

You may feel that these postulates do not tell us very much about 
points and lines. Your experience has led you to believe that there arc 
a great many points on a line and in space, and you may wonder why 
we do not say so in the postulates. We feel that it is more instructive 
to proceed as we have done, Shortly we will have introduced enough 
postulates that we will be able to prove that there are infinitely many 
points on a line and in space. However, there is not much we can prove 
from only the three postulates. Figure 1-19 should suggest to you at 
least two statements concerning lines that are not stilted in the postu- 
lates. Before formulating these statements, we will introduce some 
Symbols for points, lines, and planes. 

Notation. Capital (upper case) letters are used to denote points and 
small (lower case) letters to denote lines. For example, in Figure 1 -19 
P, Q, and R denote points and /, m, and n denote lines. We may also 
name a line by naming two points that determine the line. Thus cither 

of the symbols PQ or Qr could be used to name the line that is deter- 
mined by points P and Q. It may also be convenient to name several 
different lines by using the same letter with distinguishing subscripts 
such as h, I2, fa* * • • « 

In the figures we represent a line by the symbol 4 ► 

where the arrowheads are used to indicate that the line does not stop 
where our picture stops but continues indefinitely in both directions. 

We often use the Creek letters a, /?, v, . , , (alpha, beta, gamma, . . .) 
to name a plane, Sometimes it is convenient to name a plane by nam- 
ing three noncollinear points that the plane contains, such as "plane 
FQfi* in Figure 1-19. 

Let us return now to our discussion of the relationships between 
points and lines. Figure 1-19 suggests, but our postulates do not tell 
us, that there are at least three distinct lines in space. Let us now at- 
tempt to deduce this statement from the three postulates we already 
have. 

The Three- Point Postulate tells us that there are at least three 
points in space that do not all lie on one line. For convenience we name 
these points P t Q> and R. It follows from the Point-Line Postulate that 



1.6 Relationships of Fbints and Lines 25 

points P and Q determine exactly one line PQ, that points Q and R de- 

terminc exactly one line QR, and that points P and R determine exacdy 

one line PR. No two of these lines arc the same. For if they were, points 
P, Q, and R would be collinear, and this would contradict the fact that 
P, {), R do not all lie on one line. This completes the proof of the fol- 
lowing theorem. (Usually the proof of a theorem follows the statement 
of the theorem. For Theorem 1.1, however, the proof precedes the 
statement of the theorem.) 

THEOREM 1, 1 There are at least three distinct lines in space. 

We observe from Figure 1-19 that no two distinct lines intersect 
in more than one point We proceed now to deduce this statement 
from our postulates. 

Tf /, and t 2 are any two distinct lines that intersect, then, by defini- 
tion of intersect, we know they have at least one point in common. 
Call this point P. Now, either they have a second point in common or 
they do noL Let us suppose that they do have a second point Q in com- 
mon as suggested In Figure 1-20. 




Figure 1-20 

The Point-Line Postulate tells us that for every two distinct points 
there is one and only one line that contains them. Thus h = h (diat is, 
they are the same line). Rut this contradicts our hypothesis that k and 
h are distinct tines. Since our supposition thai l 1 and fe intersect in a 
second point leads us to a contradiction, we must conclude that h and 
I2 intersect in no more than one point. We have proved the following 
theorem. 

THEOREM 1,2 If two distinct lines intersect, then they intersect 

in exactly one point. 

Does Figure 1-20 confuse you because the marks that represent 
lines do not look as you think straight lines should look? Actually the 
word straight is not part of our formal geometry. We want a formal 



26 Points. Lines, and Plans* Chapter 1 

geometry of points, lines, and planes that is consistent with our experi- 
ences in informal geometry. In particular, we want lines in our formal 
geometry to have the property of straight ness. It is true that we cannot 
draw a convincing picture of two distinct lines with two distinct points 
in common. Tills is evidence at the informal geometry level consistent 
with Theorem 1.2. 



EXERCISES 1.8 



L Winch postulate asserts that there are at least two distinct points on 
everj' line? 

2. Restate the Line-Point Postulate and the Point-Line Postulate in your 
own words. 

3. Restate the definition of "intersect" as the word applies to two sets of 
points. 

4* If points A, B, C, and D are distinct, and if line / contains points A, B, 
and C and line in contains points A, C, and D t what can you conclude 
from this? 

5. In proving Theorem 1 . 1 wc started with three noncollincar points and 
showed that there are three distinct lines each containing two of the 
three noncollincar points. Suppose we start with four points, no three 
of them collincar. How many distinct lines are there each containing 
two of the four points? 

6. Suppose we .start with five distinct points, no three of them coBmear. 
How many distinct lines are there each containing two of the five 
points? 

7. Suppose we start with six distinct points, no three of them collincar. 
How many distinct lines are there each containing two of the six 

points? 

8. Consider your answers to Exercises 5, 6*, and 7. Can you predict how 
many lines there are for seven distinct points each containing two of 
the seven points if no three of the points are collinear? If so, how 
many? 

9. Let m and n lie different lines. Let A be a point snch that A g m and 
A <E n, Let B he a point such tliat B £ m and B C «• What must be true 
of points A and B? What postulate or theorem supports your answer? 

10. Let A and B be different points. Let line i% contain A and B and let line 
/a contain A and B, What can you conclude iihout / t and l z ? What pos- 
tulate or theorem supports your conclusion? 



1.6 R#l»tiofl ships of Points and Lines 27 



11. Consider the set consisting of the five points,, and no others, named in 
the following figure. If the points appear to be eollinear, assume that 
they are, 




(a) Identify, by listing the members, two subsets each containing three 
eollinear points. 

(b) Identify, by listing the members, four subsets each containing four 
noneollmear paints of which three points are unilinear. 

(c) Ust the members of one subset contai ning four noneollinear points 
of which no three are eollinear. 

(d) Name three lines which are not drawn in the figure, but each of 
which contains two points named in the sketch. 

12. From which postulate does it follow that no line contains all points of 
space? 

challenge pnonLUM* Start with three distinct objects such as three boys 
named Jerry, Jim, and John, Create a structure as follows: Each boy is a 
point and there is no other point. Thus space = {Jerry, Jim, John}. There 
are three distinct lines a, h, and c (and no others) as follows: a = {jerry, 
Jim}, b = {Jim. John), c = (Jerry, John}. Exercises 13-30 are questions 
about the Jerry- Jim-John structure. Kxercises 31-34 are general questions. 

13* Are the points, Jerry, Jim, and John, eollinear? 

14. Does space contain at least three noncollinear points? 

15. Does the structure satisfy the Three-Point Postulate? 

16. Is a a set of points? 

17. Does a contain at least two distinct points? 

18. Is b a set of points? 

19. Does h contain at least two distinct points? 

20. Is c a set of points? 

21. Docs c contain at least two distinct points? 

22. Does the structure satisfy the Line- Point Postulate? 

23. Is {Jerry, J tin} a set of two distinct points? 



28 Points, Lines, and Planes Chapter 1 

24, Is {Jerry, John) a set of two distinct points? 

25. Is {Jim, John} a set of two distinct points? 

26. Are there other sets of two distinct points? 

27, Is there one and only one line that contains Jim and John? 
2S, Is there one and only one line that contains John and Jerry? 
29, Is there one and only one line that contains Jerry and Jim? 
JO. Does the structure satisfy the Point- Line Postulate? 

31, Can we prove, using only the plane postulates of incidence, that space 
contains at least four distinct points? 

32, What do the Plane Postulates of Incidence tell us about the nature of 
an individual point? 

33, What do the first three postulates tell ns about the straightness of a line? 

34, What do the first three postulates tell us about segments? 



1.7 THE INCIDENCE RELATIONSHIPS 
OF POINTS, LINES, AND PLANES 

We are now ready to list our remaining Incidence Postulates. We 
begin by stating the following useful definition. 

Definition 1.3 The points of a set arc eoplanar if and only 
if there is a plane %vhich contains all of them. The points of a 
set are noncxmluiuir if and only if there is no plane which con- 
tains all of them. 



Space Postulates of Incidence 

POSTULATE 4 (The Four-Point Postulate) Space contains at 
least four nonco planar points. 

POSTULATE 5 (The Plane-Point Postulate) Every plane is a 
set of points and contains at least three noncollinear points. 

POSTULATE C (The Point-Plane Postulate) For every set of 
three noncollinear points there is one and only one plane that concur 
them, 

POSTULATE 7 (The Flat-Plane Postulate) If two distinct 
points of a line belong to a plane, then every point of the line belongs 
to that plane. 

POSTULATE 8 (The Plane-Intersection Postulate) If two dis- 
tinct planes intersect, then their intersection Is a line. 



1,7 Relationships of Points, Lines, and Planes 29 

Note that our postulates do not include a statement about the num- 
ber of planes in space. Theorem 1.3 states that there are at least two 
distinct planes in space. In the Exercises that follow you will be asked 
to prove that there are at least four distinct planes in space. 

From Postulate J we know that there arc aL least three noneollincar 
points in space. Call these points A, B, and C, By Postulate 6 there is 
exactly one plane a that contains these points. From Postulate 4 there 
is a fourth point D that is not in plane a. Points A, B t and D are not 
unilinear, for if they were, point D would lie in plane a. Why? By 
Postulate 6 again there is exactly one plane /? that contains points A, 
B, and D. Planes a and f$ are not the same plane, for if they were, points 
A, B, C. and D would be coplanar, which is contrary to the way they 
were chosen. We have proved the following theorem. 



THEORF.M L3 Space contains at least two distinct planes. 

It is often helpful to draw diagrams or pictures showing the rela- 
tionships between points, lines, and planes when making; a deduction 
such as the one for Theorem 1,3. Figure 1-21 shows points A, B, and C 
in plane a and points A, B. and D in plane /?. (Although we often use 
a quadrilateral to represent a plane, be careful not to think of the sides 
of the quadrilateral as the "edges" or "ends" of the plane.) 

Figure 1-2] also helps us to "see" that when two planes intersect, 
their intersection is a line as we have stated in Postulate 8. What is the 
intersection of a line and a plane not containing the line? When we 
look at Figure 1-22, it appears that the intersection is a single point 




L 



Figure 1-21 



1 



Figure 14S 



30 Points, Linti, and Planet Chapter 1 

THEOREM 1.4 if a line intersects a plane thai does not contain 
the line, then the intersection is a single point. 

We can prove Theorem 1 ,4 by the same technique used in proving 
Theorem 1.2. Head again the proof of Theorem 1.2 which appears 
hefore the statement of the theorem. Now try to answer the questions 
in the argument (proof) that follows* 

By hypothesis and the definition of intersect, line m and plane of 
(Figure 1-22} have at least one point F in common. How do we know 
that plane a does not contain every point of line m? Suppose line mi 
and plane a have a second point Q in common. (Here we are examining 
one of only two possibilities: either the line and plane have a second 
point in common or they do not. Of course, we are trying to prove 
that they do not.) But if a plane contains two distinct points of a line, 
then it contains the entire Line. (What postulate are we using here?) 
What does this last conclusion, namely, that the plane contains the 
entire line, contradict? Since we have agreed to accept our postulates 
as true statements al>out points, lines, and planes, what must we con- 
clude? Docs this prove the theorem? 

It follows from Postulate 6 (the Point-Plane Postulate) that three 
noncoUinear points determine exactly one plane. That is, there is 
exactly one plane which contains any set of three noncoUinear points. 

We conclude this section with two additional theorems regarding 
the existence of planes. 

THEOREM 1.5 If m is a line and P is a point not on m, then there 
is exactly one plane that contains m and P. 

Note that we must prove two tilings. 

1. There is at least one plane t hat contains m and P. 

& There is no more than one plane that contains m and P. 

These two statements illustrate the ideas of existence and uniqueness. 
When we prove existence, we show that there is at least one plane con- 
taining m and P. When we prove uniqueness, we show that there is 
at most one such plane. If we can prove both existence and unique- 
ness, then we know that there is exactly one such plane. 

In the following proof name the postulate or theorem that justifies 
the statement preceding the question "Why?". 

Proof of I. Existence. There is a plane a that contains line m and 
point P (Figure 1-23). 



1.7 Relationships of Points, lines, and Planes 3 1 

There arc at least two distinct points on line m. Why? Call these 
points A and B. The points A and B do not lie on any line except m. 
Why? By hypothesis, point P does not lie on line m. Hence points A, 
B, and V arc noncoUinear. There is a plane tt Lhat contains points A, 
B t and P. Why? If plane a contains points A and B, then it contains 
line ro. Why? Hence there is a plane that contains line m and point P, 



Mgurc 1-23 

Proof of 2. Uniqueness. There is no more than one plane tJiii 
contains line m and point P. Suppose that there is a second plane /? 
which contains m and P. Then /? contains the noncoUinear points A, B t 
and P. Thus we have two distinct planes* a and j8 s each containing 
three noncoUinear points. What postulate does this contradict? Does 
litis prove that there is no more than one plane containing m and F? 

THEOREM L6 If two distinct lines intersect, then there is 
exactly one plane that contains them. 

Let lines m and n intersect in the point P (Figure 1-2-4). 




a 



Figure 1-24 



We must prove the following: 

1. Etafatencft There is at least one plane that contains Hi and m. 

2. Uniqueness. There is no more than one plane that contains 
m- and it, 



32 Points, Lints, and Planes Chapter 1 

In the following proofs name the postulate or theorem that justifies 
the statement preceding the question "Why?" 

Proof of 1; Lines m and n have exactly one point in common. Why? 
There is a point A on m that is different from P t Why? A is not on n by 
Theorem 1.2. (Two distinct lines intersect in no more than one point.) 
There is a plane a which contains A and n. Why? Plane a contains F 
and A and therefore contains m. Why? Hence there is a plane a which 
contains m and n. 

Proof of 2; Suppose there is a second plane that contains both m 
and n. Then plane contains A because A is on m. Thus both a and fi 
contain A and n. What theorem docs this last statement contradict? 
Does this prove that there is no more than one plane a containing m 
and n? 



EXERCISES 1.7 

Refer to Figure 1-25 in working Exercise* 1-12. Assume that points A t B, 
and C are in the plane of the paper and that point D is not 




Figure 1-19 

1* Name all the planes, such as ABC, that are determined by the points 
in the figure 

2. Name atl the lines that are determined by the points in the figure. One 
of them is line A£. 

3. How many lines arc there in each plane? 

4. How many planes contain line BD? Name them. Is this number of 
planes the same for every line? 

5. How many lines are on every point? Name the lines that are on point D. 

6. How many points do plane ABC and line 55 have in common? What 
theorem justifies your conclusion? 



1.7 Relationships of Points, Lines, and Planes 33 

7. Name two planes, unci name a point that is contained in both planes. 
Which postulate tells ils that these two planes have a line in common? 
Name the line. 

8. There are at least how many points common to every pair of planes in 
the figure? 

9- Two planes which do not intersect arc said to be parallel. On the basis 
of our first eight postulates, do you think it can be proved that there 
must be two parallel planes in space? 

10, Is it possible for three planes to have one and only one point in com- 
mon? If your answer is "Yes/' name three planes and the single point 

they have in common. 

11, Two noncoplanar lines in space are said to be skew lines. Name three 
pairs of lines in the figure such that each pair is a pair of skew lines. Can 
a pair of skew lines intersect? Why? 

12. Can three lines have a single point in common and be noncoplanar 
lines? 

13, Let P be a point in plane a. On the basis of only our first eight postu- 
lates do you think that we can conclude that there are three distinct 
Hues in plane a that contain point F? Give reasons for your answer. 

14. How many points do the planes of the ceiling, front wall, and side wall 
of your classroom have in common? On the basis of only our first eight 
postulates do you think it can be proved tint there arc three distinct 
planes that have more than one point in common? 

15. Using three pages of your book as physical examples of planes, what 
would you guess to be the intersection of three distinct planes if then- 
intersection contained more tlian one point? 

16\ I Ay the edge of your ruler along the top of your desk. Do all the points 
on the edge of the ruler seem to touch the desk? Which of our postulates 
docs this illustrate? 

In Exercises 17-25, write a short paragraph to show how our postulates or 
theorems can be used to prove the .statement. 

17. Every line is contained in at least one piano. 

18. There are at least three distinct lines in every plane. 

19. There are at least four distinct planes in space. 

20. There are at least six distinct lines in space. 

21. If four points are noncoplanar, then they are noneollinear. 

22. If four points are noncoplanar, then any three of these points are 
noneollinear. 

23. There are at least three distinct lines through every point. 

24. Every line is contained in at least two distinct planes. 

25. Every point is contained in at least three distinct planes. 



34 Points, Ones, and Planes Chapter i 

26. challenge problem. Suppose, for this one exercise, that we replace 
our Postulate 8 with this postulate: "If two distinct planes have one 
point in common, then they have a second point in common." Prove 
our replaced Postulate 8 as a theorem. 

27. challenge problem. Suppose, just lor Exercises 27 and 28, that 
space consists of four distinct points A, B, C, D and no others. Suppose 
that there is one line a containing all four of the points A, B, C, D and 
that there are no other linos. Suppose that there are no planes. Does 
this structure satisfy Postulate 4? Since A, B, C, and D are not contained 
m any plane, it follows that they arc noncoplanar, and hence Postulate 
4 is satisfied. 

Does this structure satisfy Postulate 5? Yes, it does. To show lliis 
we must check each plane in the structure to see if it contains three 
noncollinear points. If there is no plane that violates the property of 
Postulate 5, then Postulate 5 is satisfied. Since there is no plane in the 
structure, it follows that there is no plane that violates Postulate 5, and 
hence Postulate 5 is satisfied. Show that Postulates 6, 7, and 8 arc satis- 
fied by this structure, 

28. challenge p itonLEM . Does the structure of Exercise 2 7 satisfy Postu- 
late 1? Postulate 2? Postulate 3? Your intuition might tell you that Pos- 
tulate 4 includes Postulate 1 in the sense that Postulate 1 could he 
proved once Posulate 4 is assumed. The structure of Exercise 27 pro- 
vides an answer to the question; Is it possible to satisfy Postulate 4, in- 
deed all of the postulate, from 2 through 8, without satisfying Postulate 
1? In other words, is Postulate 1 independent of Postulates 2 through 8? 
Why? 



CHAPTER SUMMARY 

In tliis first chapter we compared informal geometry and formal ge- 
ometry noting that geometry began informally many years ago as a collec- 
tion of rules to solve practical problems related to "earth measurement." 
Although geometry as a school subject is both formal and informal, the ap- 
proach in tins book is predominantly formal. 

Tile formal approach features carefully stated postulates, definitions, 
and theorems. Proving theorems involves- deductive reasoning. We dis- 
cussed examples of intuitive reasoning, inductive reasoning, and deductive 
reasoning. 

We reviewed the language of sets. If A and B are sets ? the INTERSEC- 
TION of A and B, denoted by A n B t is the set of elements contained both 
in A and in B, The UNION of A and £, denoted by A U B, is the set of ele- 
ments contained in A or B, The intersection of two sets may be die 
EM PTY or NULL SET denoted by . If two sets INTERSECT, they must 
have at least one element in common. 



Review Etercis** 35 

If p and q arc statements; the CONJUNCTION of these two state- 
ments is the statement p and q. The conjunction of p and o; is true if both p 
and ^ arc true, but false in all other cases. The DISJUNCTION of the two 
statements is the statement porq. The disjunction of p and q is false if both 
p and q are false, but true in all other cases* 

In order to avoid circular definitions w« accepted POINT, LINE, and 
PLANE as UNDEFINED TERMS in our formal geometry. 

We defined SPACE to be the set of all points. The points of a set are 
COLLINEAR if and only if there is a line that contains all of them. The 
points of a set are COPLANAR if and only if there is a plane that contains 
all of them. 

Our first eight postulates are called INCIDENCE postulates and arc 
listed below by name only. Be sure that you can state each of them in your 
own words. 

1. THE THREE-POINT POSTULATE 

2. THE LINK-POINT POSTULATE 

3. THE POINT-LINE POSTULATE 

4. THE FOUR-POINT POSTULATE 

5. THE PLANE-POINT POSTULATE 

6. THE POINT-PLANE POSTULATE 

7. THE FLAT-PLANE POSTULATE 

8. THE PLAxNE-INTEftSECnON POSTULATE 

In addition to these eight postulates we stated and proved six theorems. 
Read these theorems again and study their proofs. 



REVIEW EXERCISES 

In Section L I an example involving the area of a right triangle was given to 
illustrate intuition or iutuitive reasoning. In Exercises 1-10, state whether 
or not you think the given example is a good example of intuitive reasoning. 
If you think the decision made is false, indicate it. 

1. Examine a rectangular box and decide that the opposite faces of a rec- 
tangular solid have the same size and .shape. 

2. Examine a ball and decide that a plane through the center of a spherical 
region (a Sphere and its interior) would separate it into two hemispheri- 
cal regions with equal volumes. 

3. Examine a ball and decide that the area of a sphere is four times the 
area of a circular cross section formed by a plane passing through the 
center of the sphere. 

4. Examine three distinct points on a line and decide that if A, B y C are any 
three distinct points of aline with B between A and C, then the distance 
from A to H is less than the distance from A to C. 



36 Points, Unas, and Pfants 



Chapter 1 



5. Think about a triangle Trying in a plane a. Think about a line in or that 
passes through a point F lying inside of T. Decide that this line must 
contain two distinct points of T. 

6*. Think about a triangle, a quadri- 
lateral, and a pentagon. Decide that 
the sum of the measures of the 
angles is the same for each. 

7. Think about three books of 300 
pages each standing upright on a 
shelf as suggested in the figure. De- 
cide that it is farther from page 1 of 
book I to page 300 of hook 1 II than 
it is from page 300 of book 1 to page 
1 of book III. 

8, Think about a plane figure consisting of a triangle and the bisector of 
one of its angles as suggested in the following figure. Decide that the 
bisector of an angle of a triangle divides the opposite side into two seg- 
ments of equal length. 





9. Tliink about a circle and a line that separates it into two arcs of equal 
length. Decide that the line passes through the center of the circle, 

10. Draw a triangle Mark one point on each side as close to the midpoint 
as you can without measuring. These points are vertices of a second 
triangle. Decide thai the area of the original triangle is four times the 
area of the second triangle. 

In Section LI an example involving the sum of the measures of the angles 
of a triangle was given to Illustrate inductive reasoning. In Exercises 11-20 
draw one or more figures and make some measurements if you wish. Then 
draw an appropriate conclusion. 

11. Is the following statement true or false? The bisectors of the angles of 
a triangle are concurrent, thai is, they all pass through the same point. 

12. What is the sum of the measures of tlie angles of a pentagon? (Consider 
only pentagons whose diagonals pass through the interior of the 
pentagons.) 

13. If A BCD is a quadrilateral with AB = BC, h it necessarily true that 
CD = DA? 



Review Exercises 37 

14. If ABCD is a quadrilateral with AC = HD t is it necessarily true that 
AB = BC? 

15. If the sides of a quadrilateral have equal lengths, is it necessarily true 
that its diagonals have equal lengths? 

16. If the diagonals of a quadrilateral bisect each other, is the quadrilateral 
necessarily a parallelogram? 

17. If the sides of AABC have equal lengths, if D is the midpoint of BC, 
If E is on line DA, if D is between A and E» and if AD — DE, is it true 
that S?is parallel to AC? 

18. If AABC is any triangle and if D is the midpoint of EC* is it necessarily 
true that AABD and A A CD have equal areas? 

19. If AABC is a triangle and if D is the midpoint of B~C, is it true that 

AD = %AB + AQ? 

20. If A ABC" is a scalene triangle (no two of its sides have the same length), 
is it true that no two of its angles have equal measures? 

Section 1,2 contains two examples of deductive reasoning. In Exercises 21- 
25, state whether you think the conclusion C (Ci, Ca, . . - , if more than one) 

follows logically from the hypothesis H (Hi, 91% if more than one). Be 

prepared to explain what assumptions you are making, if any, in addition 
to those that are stated as hypotheses, 

21. H. x = o 

C: 2* -f- 3 = 13 

22. Hi 2x + 3 = 13 
Ct x = 5 

23. Hi'. Every polygon is a plane figure, 
ll'i- P is a polygon. 

& P is a plane figure. 

24. B Let AABC and ADEF be given. If ml A = mLD, mlB = 

B»Z£, and AB = DE, then AC = DFsad BC = EF. 
G; tx;t AABD be given with C between A and D. If m / ABC = 

ml VBC and mlACB = m I DCB, then AC = CD 
Q: Let AABC and ADEF be given. If ml A ~ mZD, AB = 

DE f and AC = DE, then BC = EF. 

25. Hi: ATI cows eat grass. 
H& Bessy is not a cow. 

Ct Bessy does not eat grass, 

26. If A is the set of even integers and B is the set of odd integers, describe 
the union of A and B< the intersection of A and B. Do sets A and B 
Intersect? 



38 Points, Lines, and Plane* Chapter 1 

■ Exercises 27-38 refer to the three sets 

Q= {-3, 0, 3 r G), 

<J= {-3, 0, 3, 6,9}, 

T = [x \ x is an Integer, - 3 <; x < 9, and x is divisible by 3), 

27. Is Q C S? 33. Is T C S? 

28. Is S C 9? 34. Docs S = 7? 

29. Is Q c 7? 35. Find Q D S. 

30. Is T C 0? 3fl - Uoe* D S = Q? 

31. Does Q = T? 37. Find Q U S, 

32. Is S c T? 38. Does £ U S = S? 

39. If A = {x ; x £ R and x < 7} and B = (x : x £ /I and x > 3}, draw 
the graphs of A, B, and A H Bon three separate number lines. 

40. Describe the set A H /J in Exercise 39 using set-builder notation. 

41. If S = {x : x £ R and x > 5} and T = {x : x C R and x < 0), draw 
the graphs of S, T, and S U Ton three separate number lines. 

42. Describe the set S U T in Exercise 41 using set-builder notation. 

■ If p is the statement 7 = 7, q is the statement \/9 as 5, and r is the statement 
— 5 < —2, determine in Exercises 43-49 if the given "compound" state- 
ment is true or false. 

43. p and q 47. q and r 

44. p or q 48. q or r 

45. n and r 49. (p and q) or r 

46. por r 

50, State the Flat- Plane Postulate in your own words. 

51. The following figure is a picture of two distinct "curves" that intersect 
in two distinct points. State a postulate or theorem that would settle an 
argument about whether these " curves'" might be lines in our formal 
geometry. 




52. Iff and Q are distinct points and m and n are lines, and if P £ m, Q € ™> 
P E ?t, and Q £ n, what conclusion can you draw? What postulate or 
theorem justifies this conclusion? 

53. If m and n ai'e distinct lines and P and Q are points, and if m fl n = F 
and m n n — Q, what can you conclude? What postulate or theorem 
justifies this conclusion? 



Review Exercises 39 

54. Which postulate assures us that no plane contains all the points of 
space? 

55. Prove, using only the postulates, that every point is contained in at least 
one plane, 

56. Draw a single diagram which illustrates what all eight postulates assert. 
Label the points in your diagram and then name one pair of skew lines, 

57. What dees it mean to say that points A t B, and (7 are collinear? 

5S, Does the Three-Point Postulate eliminate the possibility of there being 

tliree points in a plane that are collinear? 

59. Assume that A, B, C, D, and E are five distinct points in the same plane 
with no three of these points collinear. 

(a) How many distinct lines are there each containing two of the five 
points? 

(b) How many lines determined by A, B, C t D, and E are there on each 
point? 

60. Let A, J3, C, D, and E he five distinct points in the same plane. 

(a) If A, B, and C are on line 1, and C. D* and F are on line m, and l^m, 
how many more lines are there each containing two of the five 
points? How many lines are there altogether? 

(b) How many distinct lines pass through point A? Point #? Point C? 

6L If line I contains three distinct points A, B, and C and if points A and B 

lie in plane a, what can you conclude? 
62. Planes RST and S7T- are distinct. 

(a) Name a line that lies in both of these planes. 

(b) Name two more planes determined by points H. S. 7", U. 
6,1, tState what intersect means 

(a) with regard to two lines, 

(b) with regard to two planes, 

64. Which of our Incidence Postulates "pushes" us off a plane into space? 

Determine in Exercises 65-71 if the given statement is true or false. 

65. It is a theorem that every line contains at least two distinct points. 

66. The Line- Point Postulate assures us that there is just one line which 
contains two distinct points. 

67* It is a theorem that there arc at least throe distinct lines in every plane, 

68. If a line intersects a plane not containing the line, the intersection is a 
single point 

69. It is a theorem that there are at least four distinct planes in space. 

70. Our Incidence Postulates assure us of at least twelve distinct lines in 
space. 

71. The Three-Point Postulate assures us that there are at least three dis- 
tinct points on every line. 





SydCnrnberg/lXF.t. 



Separation 
and Related 
Concepts 



2.1 INTRODUCTION 

In this chapter we list our second group of postulates. These postu- 
lates are concerned with points between other points, how a point sep- 
arates a line, how a line separates a plane, and how a plane separates 
Space, The ideas these postulates convey arc very simple ones, and the 
information they give can be easily determined by looking at pictures. 

We list these postulates so that we can give a meaning to the rela- 
tionships of bctweenness and separation and to make it quite dear 
what statements we agree to use in our proof & Later in the chapter we 
use these relationships to make several definitions. 

2.2 THE BETWEENNESS POSTULATES 

In this section we state the Betwecnness Postulates. They are some- 
times called order postulates because they tell us how points are ar- 
ranged in order on a line. For example, if A* B, C are three distinct 
points on aline, and if Bis between A and C, then we could say that the 
points are arranged on the line in the Order A* then B, then C We could 
also say that the order is C, then B t then A, (See Figure 2-1.) 



a b C 

-• • 9- 



Frgrirc 2-1 



42 Separation and Related Concepts 



Chapter 2 



In our formal geometry between and order, as they apply to points, 
are undefined terms. We do not explain by a definition what it means 
to say that point B is between points A and C Rather we accept the 
following Betweenncss Postulates as formal statements of betweenness 
properties. 



The Betweenness Postulates 

POSTULATE 9 (The A-B-C Betweenness Postulate) If point 
B is between points A and C, then point B 1% also between C and A, 
and all three points are distinct and collinear. 

Note that the postulate does not allow us to say that "point B is 
between points A and C" if the points are as shown in Figure 2-2a. 






Ftpir* 2*1 

Why? Similarly, in Figure 2-2b point B is not between points A and C. 
Why? Figure 2-2c is a correct picture of what we mean when we say 
"point B is between points A and C." We use the symbol A-B-C or 
the symbol C-B-A when we wish to say that "point B is between points 
A and C." 

POSTULATE 10 (The Three-Point Betweenncss Postulate) U 
three distinct points are coUincar, then one and only one is between 
the other two. 

From this postulate it follows that if A, B t Care three distinct points 
on a line, then B-A-Q or A-B-C, or B-C-A\ and if A-B-C, for exam- 
ple, then we cannot have B-A-C or B-C-A, From this postulate it also 



2,2 The Betweenness Postulates 43 

follows that betweenness for points in our format geometry is different 
from betweenness on a circle. For example, if three people are seated 
at a circular table, then most of us would agree that each of them is 
between the other two. But we should also now agree, after accepting 

Postulate 10, that betweenness around a table is different from be- 
tweenness in our formal geometry. 

POSTULATE 11 {The Line-Building Postulate) If A and B are 
any two distinct points, then there is | point Xj such that X a is between 
points A and B, a point Yg Sttdi that B is between A and Y lt and a point 
Zi such that A is between Zj and B. 

Figure 2*3 illustrates the meaning of Postulate 1 L We call tliis pos- 
tulate the Line-Building Postulate because it enables us to prove that 
there are infinitely many points on a line. 

Zi a X x B Yi 
4 * * • • * ^ 

Pigut* £-3 

From this postulate it follows that there is a point Yi on line AB 
that is beyond B from A, a point Y 2 that is beyond Y x from B, a point 
Y 3 that is beyond Yt from Y\, and so on (Figure 2-4). 

Z\ A Xj B y, F 2 Yi Y± 

4 • • • • • * — # — • » » « — ► 

He..«v v. i 

Tlius we are able to find on the line as many points as we choose, 
3 or 30 or 3,000,000 or any natural numl>er you wish, that arc beyond 
B from A. Similarly, we can find as many points as we choose that are 
beyond A from B (Figure 2-5). 

Z4 ?a Zi Z\ A Xi B Yt Y 2 Y 3 Y. t 

4 — • • • • — • — • • • • • • • — • — • — • — •— . ■ ^ 



Figure 2-5 

From this postulate it follows also that no matter how close 
together two distinct points may be there is always a point between 
them (Figure 2-6). 

Z% Z-i Zj A Jfj. X 2 Xs X* Xn B Yi Yi Y S 
i •• • • • • — * — • • • • » * • • — • m » # | 

Figure £-6 



44 Separation and Related Concepts Chapter 2 

Does your intuition tell you that Postulate 11 builds every line com- 
pletely, that no "holes" could possibly exist in a line? Actually, Postu- 
late 11 assures 08 that there arc infinitely many points on every line and 
furthermore, infinitely in any points between every two points of a line. 
But it does not assure us that there arc no holes in a line. The postula- 
tional basis for lines is completed in Chapter 3, where we adopt a Ruler 
Postulate. This postulate provides for the ideas of "continuity' 1 and 
"infinite extent" 



EXERCISES 2,2 

In Exercises 1-22, indicate by 1" or F whether or not the statement is true 
or false. 

In Exercises 1-4, R, S, Tare pointi such that ft-S-T. 

1. /i„ $ ? T are collinear points. 

2. R. 5, Tare distinct points. 

3. T-S-R 
4 S-R-T 



In Exercises 5-KJ, points A and B are on line m-, point C is between A and 
H and on line n. and m and n arc distinct Hues, 

5. m and n intersect at A. 

G. m and n intersect at B> 

7. m and n intersect at C 

8. A, B, C are collinear points. 

9. A-B-C 
10. C-B-A 



hi Exercises 11—18, F, Q, R are distinct points and no one of them is between 
the other two. 

11. F, Q, R are collinear points. 

12. P and Q are collinear points. 

13. P and R are collinear points. 
14 Q and R are collinear points. 

15. There is one and only one plane that contains P, Q, and R, 



2 J> The Betweenness Postulates 



45 



16. There is a point S such that Q-R-S. 

17. There Is a point 7 such that Q-T-H, 
IS. There is a point U such that U-Q-R, 

In Exercises 10-223 R, S, Tare three distinct eollinear points, R-S-T is false, 
and 5-R-7" is false, 

19. R-T-S 

20. S-T-R 
2L T-R-S 
22. T-S-R 

Exercises 23-32 refer to Figure 2-7 which shows three noncolhnear points 
A, B, C and the three lines and the one plane that they determine. 




Figure 2>7 

23. Which Incidence Postulate asserts the existence of three noncoUinear 
points? 

24. Which Incidence Postulate assures us, once wo have points A, B t C, 
dial lines AB t BC, CA exist? 

25. Which Incidence Postulate assures us, once we have points A, B, G\ 
that there is a plane or containing them? 

26. Which postulate assures us. once we have the points and the lines of 
the figure, that there are points D t Z, F such that B-U-C % C-E-A, 
A-F-H? 

27. Explain why D and E are distinct points. 

28. Why is D-E-A a false statement? 

g& Why is E-A-F a false statement? 

30, Which postulate assures us that diere are points G and H such that 
E-P-C and D-F-IT? 

31, Explain why // and C are distinct points. 

32, Explain why AB and HG are distinct lines. 



46 Separation and Related Concept* Chapter 2 

■ Exercises 33-35. The Incidence Postulates assert the existence of four non- 
coplanar points. Figure 2-6 suggests four noncoplanar points A, B, C„ D and 
the four planes and sis lines determined by them. 




Ffgun»4 

33. Deduce from the Incidence and Betweenness Postulates that there is 
at least one point in space that is not contained in any of the four planes 
of the figure* 

34. Deduce from tl»e postulates that there is at least one line in space that 
is not contained in any of the four planes of the figure. 

35. Deduce from the postulates that there is at least one plane in space that 
is different from any of the four planes of the figure- 
Exercises 36-42 refer to Figure 2-9 which shows a line and a portion of it 
(the heavier part including the points C and D) called a segment (Segment 
is defined in the next section,} Points X, Y, 2 as well as C and D are points 
of this line as indicated in the figure. 

< I £ 1 — I i ► 

Figure 2-9 

36. Are X and V points of the segment? 

37. Is Z a point of the segmont? 

38. Is C-X-D? 40. Is C-Z-D? 

39. Is C-Y-D? 41. Is C C m 

42. Try to write a definition of segment. 

43. challenge pboblem. Let a line I be given. Deduce from the postu- 
lates that there are at least three different planes containing /. 

44. challenge problem. Let P be u point in a plane a. Deduce from the 

postulates that there are at least three dLUereuL lines in a passing 
through P. 



2.3 Using Batweenne&s to Make Definitions 47 



2.3 USING BETWEENNESS TO MAKE DEFINITIONS 

We now list several definitions that make use of the between ness 
relation. Most of these are definitions of terms you have met in your 
earlier work in geometry. However, it is the betweenness relation that 
enables us now to make these definitions precise. 

Figure 2-10 is a pic Lure of a set of points called a segment 

A B 
»■ ■ • 



Figure 2-10 



Definition 2,1 If A and B are any two distinct points, seg- 
ment AS is the set consisting of points A, B, and all points 
between A and B. The points A and B are called endpoints 
of AS 



Sometimes a segment is called a line segment to avoid possible con- 
fusion with a "segment" of a circle. Be careful not to confuse the sym- 

bolAi? for a segment with the symbol AB for a line. The horizontal bar 

in AB reminds us that a segment has endpoints, whereas the bar with 

* — * 
arrows in AB reminds us that a line has no endpoints. 

We sometimes say that two distinct points determine a segment, 

T hus, if A and B are two distinct points, then the segment determined 

is AB. Similarly, if A,_B, C are three distinct points, then the segments 

determined arc AB, BC, and CA. 



1 

-#- 



Figure 2- II 

Figure 2-1 1 is a picture of a set of points called a ray. As the figure 
indicates, a ray has just the one endpoint A. Speaking informally we 
might say that the ray continues without end from A through B in a 
straight line. We express the idea of a ray formally in our next 
definition. 



Definition 2.2 If A and B are any two distinct points, ray 
AM is the union of segment AB and all points X such that 
A-B-X. The point A is called the endpoint of AB. 



48 Separation and Related Concepts Chapter 2 

In the symbol AB t the arrow reminds us that it is a ray. The arrow 
in this symbol is always drawn from left to right regardless of the di- 
rection in which the ray points. However, do not confuse the symbol 

AB with the symbol BA. Both represent rays, but they do not represent 

the same ray, as seen in Figure 2-12, Observe that ray AB has endpoint 

A, whereas ray BA has endpoint B, 



A B 

• •- 



VtayAB 



R*y 



A 
tytiX 



Figure 2-12 



It is often convenient to speak about opposite rays. Figure 2-13 
is a picture of what we mean by opposite rays. The figure suggests the 
following; definition. 



AC AB 



B 



Figure £-13 



Definition 2.3 If A is between B and C, then rays AB and 
AC are called opposite ruys. 



The betweenness relation C-A-B in this definition requires oppo- 
site rays to be collincar. Why? They mast also have a common end- 
point. Thus the two rays shown in Figure 2-14 are not opposite rays 
even though they "point in opposite directions." 



Q 
— 



Figure 2-14 



2.3 Using Betweenness to Make Definitions 49 

Note that the definition of opposite rays is equivalent to saying that 
two distinct rays are opposite rays if they are colfinear and have the 
same endpoint. 

We often use the symbol "opp.AB" for the ray opposite AB. Thus, 

in Figure 2-13, opp AB = AC since both symbols name the same set 

of points, namely , the ray that is opposite AB. 

The following summary should be helpful: If A and B are any two 
distinct points on a line, they determine the six subsets of the line 
shown in Figure 2-15, with subsets indicated by heavier marking. Be 

sure to notice the differences among the symbols AB, AB, AB, and BA. 



line AB = BA 4 £ fc 



segment a_ 



B 



ray AB 4 4 2- 



opp AB 
myBA 



A 



oppBA 4— ♦ 



Figure i-15 



A segment has two endpoints and a ray has one endpoint. Some- 
times we wish to speak about those points of a segment or ray that are 
not endpoints. Accordingly, we state the following definition . 



Definition 2.4 The interior of a segment is the set of all 
points of the segment except its endpoints. The interior of a 
ray, also called a halftone, is the set of all points of the ray 
except its endpoint. 



50 Separation and Related Concepti Chapter 2 

EXERCISES 2.3 

L Use the A-B-C Betwecnticss Postulate to prove that every point of AB 

is in AH. 
2. Consider the following definition of a segment AB: 

AB = (X : ,V £ AH. and X = A, or X = B, or A-X-B). 

Is this definition equivalent to Definition 2.1? 

3- Consider the following definition of a ray AB: If A and H are distinct 

points, then ray AB consists of point A, point B> all points X such that 
A-X-B, and all points V such that A-B- X Is this definition equivalent 
to (Definition 2,2? 

4. If A and B are distinct points, what is the intersection of rays AB and 
BA? 

5. If A and B arc distinct points, what is the union of AB and BA? 

6. Write an alternate definition of segment in terms of intersecting rays. 

7. Can two distinct rays have no print in common? Illustrate with a figure. 

8. Can two distinct rays have exactly one point in common? Illustrate. 

9. Can two distinct rays have infinitely many points in common? Illustrate. 

10. Can the intersection of two distinct rays be a set containing two and 
only two distinct points? 

11. Can the intersection of two distinct rays be a ray? 

12. Can two collinear rays have no point in common? 

13. Is there a segment with no endpoint? 

14. Is there a ray with no endpoint? 

15. Is there a segment with no interior point? 

16. Is there a ray with no interior point? 

17> Is there a segment with exactly one interior point? 

18. Is there a ray with exactly one interior point? 

19. Is there a segment with exactly two interior points? 

20. Is there a ray with exactly two interior points? 
2L IF A-B-C, what is the intersection of AH and AC? 
22. If A-B-C, what is the union of AB and AC? 

2a (f A-B-C, what is the intersection of AB and BC? 

24. If A-B-C, what is the union of "KB and B~C? 

25. If A-B-C, what is the intersection of AB* and ^B? 

26. If A-B-C, what is the union of AB and ~C$? 

27. If A-B-C. if S is the Interior of BA t and if T is the interior of BC. what 
is the intersection of S and 7"? 



2.3 Using Betweenness to Mak# Definitions 51 

28. If A-B-C, if S is the interior of BA, and if T is the interior of BC, de- 
scribe the union of S and T. 

29. If 5 is the interior of a segment AB, what is the union of S and ABP 

30. If S is the interior of a segment AB, what is the intersection of S and AS? 

31. If A and B are distinct points of the segment HS, what is the union of 
A3 and 55? 



In Exercises 32-41 , a subset of a Hne / is named. Points on line I are indicated 
in Figure 2-16. Using tills figure, write a simpler name for the set. Exercise 

32 has been worked as a sample. 



t 2 


5 


C 


fl 


£ J? 






















Figure 2-1 


32, CE H CE = CE 








37. tf? n *S 








.13, Ci U CE 








38. ED U CF 








34, Dp fl DA 








39. imncF 








35. o/^j DF H CF 

36, ~CF U 5? 








40. offi £? n 

41. Mnopp 


BE 


CA 





42. Prove that there is no segment on a line / which contains every point 
of line /. (Hint: Let A~S be any segment on line I Show that there is a 
point on line I that is not in AB,} 

43. Iff and Q ore two distinct points on a line, does QP = opp PQ? Explain. 

44. How many segments do three distinct points on a line determine? Four 
distinct points? Five distinct points? (A given set of points determines 
a segment if the endpoints of the segment are in the given set.) 

45. Use your answers to Exercise 44 to predict how many segments six dis- 
tinct points on a lino determine. Test your prediction by counting. 

46. (a} A, B f C t D are four distinct points on a line /. The conjunction of the 

two sentences (1) and (2) is true if the appropriate symbol ( — , -^, 
«-*>) is supplied over each letter pair (BA). Copy these statements 
and supply these symbols. 

(1) BA contains points C and D, but BA contains neither of these 
points. 

(2) D belongs to BA, but C does not. 

(b) Draw a sketch which shows the order of the four points A t B^QD 
on I. 

47. A % B, C. D are four distinct points on a line /. If C g DB, A £ DB, 
C £ Bu. A £ BA draw a sketch which shows the order of the four 
points on /.. 



52 Separation and Related Concepts 



0-,m:.:-2 



48, If FQ is opposite to PR* which one of the three points P, Q t R is between 
the other two? 

40. If opp RQ = RF, which one of the points P, Q, R is between the other 
two? 



2.4 THE CONCEPT OF AN ANGLE 

The concept of an angle is a fundamental one in mathematics as 
well as in the practical world of the house builder and the engineer. 
There are several ways to think of an angle. One way is to think of an 
angle as two noncollinear rays that have the same endpoint. A second 
way is to think of an angle as the points on two noncollinear rays that 
have the same endpoint. In one, an angle is a set of rays. In the second, 
an angle is a set of points* In the following formal definition we agree 
to think of an angle as a set of points. 



Definition 2.5 An angle is the union of two noncollinear 
rays with the same endpoint. Each of the two rays is called a 
side of the angle. The common endpoint of the two rays is 
called the vertex of the angle. 



Notation. The angle formed by the rays AB and AC is denoted by 
Z BAG. When using a symbol such as Z BAG it is important that the 

middle letter denote the vertex of the angle. Since the union of A R and 

AC is the same as the union of AC and AB, it should be clear that 
Z BAC = Z CAB, (See Figure 2-17.) It should also be clear that many 
different points may be used in identifying the same angle. Thus, if 

AB = aS=AD and AE = A? = aS, 

as indicated in Figure 2-18, then ZBAE = Z CAE = Z DAE = 
Z RAF = Z BAG, and so on. 





Figure 2-1" 



Figure £-18 



If a figure or other information makes clear which rays are the sides 
of the angle, we can denote Z BAE by simply writing Z A, However, 



2.4 The Concept of in Angle 53 

if there is more than one angle with vertex A as in Figure 2-19, we 
would not know which angle is referred to by L A, 




Figure M 



It is often convenient to label an angle by indicating a numeral or 
a lower case letter in its interior. Thus in Figure 2-19 we can write L 1 
for Z DAC and Z r for Z BAD. When used, the arcs in the figure indi- 
cate the sides of the angle and the interior of the angle. 

Sometimes we speak of the angle determined by two noncollinear 
segments which have a common endpoint as indicated in Figure 2*20. 
If A B and BC are the segments, then the unique angle determined by 
them is ZABC 




Figure %m 



Observe in the definition of an angle that the sides of an angle are 
noncollinear rays and therefore distinct rays. In some books this re- 
striction is not made, A special case of an angle results when iLs two 
sides are "coincident." It is called the /em angle. Another special case 
results if the two sides are distinct but coUinear. In this case the sides 
of the angle are opposite rays and the angle is called a straight angle. 
In Figure 2-21, Z ABC is a zero angle and ZDEF is a straight angle. 



A 




Figure Ml 



54 Separation and Related Concept! Chapter 2 

However, because zero angles and straight angles are not needed and 
it is simpler not to consider them, they have been excluded from the 
formal definitions in this book. 



EXERCISES 2,4 

1. Copy and complete the following definition: An angle is the [T] of two 
[7] which have a common ondpoinl but do not lie on the same \t]. 

In Exercises 2-9, draw a picture to illustrate a set satisfying each of the given 
descriptions. 

2. Two distinct coplanar rays whose union is not an angle. 

3. Two angles with the same vertex whose intersection is a ray. 

4. Two angles with different vortices whose intersection is a ray. 

5. Two angles whose intersection is a segment 

6. Two angles whose intersection is a set consisting of cxacdy one point. 

7. Two angles whose intersection is a set consisting of exactly two points. 

8. Two angles whose intersection is a set consisting of exactly three points. 

9. Two angles whose intersection is a set consisting of exactly four points. 

10. If A, B t C are three distinct points on AB t if A, D„ E are three distinct 

points on AD, and if A, B, D are noncollinear points, then the union ol 

18 and -A D is an angle which we may indicate as L BAD. Using A, B, 
C, D, E, write the other possible symbols for this angle. 

11. In the following figure , the capital letters denote points and the lower 
case letters demote angles. Using three capital letters, write another 
name for each angle denoted by a lower case letter. 

(a)*=ZQ (d)«=Z[?3 

(b) y= im (e) r,= Z[?] 

(c) %= ZQ (f) w= L\T\ 




2.4 Th* Concept of an Angia 55 

12. How many angles are determined by three distinct coplanar rays having 
a common endpoint if no two of the rays are opposite rays? By four dis- 
tinct rays? By five distinct rays? (A given set of rays determines an angle 
if the sides of the angle are elements of the given set.) 

13. Would your answers to Exercise 12 be different if the rays were not all 
in the same plane? 

14. In the figure below, AB and AC are opposite rays. How many angles do 
the four rays determine? Name them. 




15. If two distinct lines intersect, how many angles are determined? (A 
given set of lines determines an angle if each side of the angle is con- 
tained in one of the lines of the given set.) 

16. If three distinct coplanar lines; intersect in a common point, how many 

angles do they determine? 

17. Name all the angles determined by the segments shown in the figure. 
How many are there? Which angles can Ijc named by using only the 
vertex letter? 




18. Using letters in the figure in Exercise 17, name the 

(a) angle with vertex P in four different ways. 

(b) ray that is the intersection of APRS and I QRS. 

(c) ray that is the intersection of APSR and / QSR. 

(d) segment that is contained in the intersection of Z PQR and Z SRQ, 

(e) point that is not in FR but is in the intersection of I.RPQ and 
APRS. 

(f) three distinct rays contained in the union of Z PRS and Z QRS. 



56 Separation and Related Concept* 



Chapter 2 



Exercises 19-30 are informal geometry exercises. The formal development 
of angle measure appears in later sections. Using Figure 2-22, which shows 
degrees on a protractor, compute the number named in the exercise. Exer- 
cise 19 has been worked as a sample. 




figure 2*2a 

19. mlDAF = 55 

20. ml BAD 
21 mlDAH 
22. mlHAE 



23. mil AH 

24. mllAE 

25. mlEAD 
2«. ml IAD 



27. mllAE + tnlEAD 

28. mlDAH - mlllAE 

29. ml DAE 

30. ml DAE + mlDAF 



31. Eveiy triangle is the union of three segments, but not every union of 
throe segments is a triangle. Write a definition of a triangle that you 
think will "hold water." (The formal definition of triangle appeals in a 
later section.) 



2.5 THE SEPARATION POSTULATES 

Tn this section we state three more postulates involving die idea of 
betweenness, which we call the Separation Postulates. These postulates 
tell us how a point separates a line, how r a line separates a plane, and 
how a plane separates space. Although the ideas are very simple, we 
cannot prove them from the postulates thus far agreed on. In order to 
facilitate their phrasing, wc introduce the idea of a convex set. 



Definition 2.6 A set of points is called convex, if for every 
two points F and Q in the set, the entire segment F^Jis in the 
set. The null set and every set that contains only one point 
are also called convex sets. 



2.5 The Separation Postulates 57 

The definition implies that if P and Q are any two points of a con- 
vex set T, then PQ C T. A line, a ray* and a segment are examples of 
convex sets of points as Figure 2-23 suggests. Is a plane a convex set of 
points? 



= / + 



+1 



_ — *A 
FQt=ABm- 



PQ^Alti- 



Q 



Figure £-23 

The interior of a triangle is a convex set Figure 2-24 shows two 
choices. Pi, P* and Q r , Q 2y for points P and 9, respectively, in the in- 
terior of the triangle and in each case the entire segment PQ is con- 
tained in the set Is the interior of the circle shown in Figure 2-24 a 
convex set? (We define interior of triangle, circle, and interior of circle 
formally in later sections,) 





Fisure 2-24 



However, none of the sets shown in Figure 2-25 is a convex set In 
each instance it is possible to find points Pand Q such that not all of the 
segment PQ is contained in the set. 





J 


1 

p 

/ 



Figure 2-25 



58 Separation and Related Concepts Chapter 2 

Let S be a convex set, T a convex set, and R the intersection of the 
two sets. If R is the null set or if ft consists of a single point, why is R a 
convex set? Suppose, then, that R consists of more than one point. Let 
P and Q be any two distinct points in H. Draw an appropriate figure 
to suggest the sets S T % R and the points P and Q, Then answer the fol- 
lowing questions. 

1. Why is P in the set S? In the set T? 

2. Why is pin the set S? In the set T? 

3. Why is P£ in the set S? In the set T? 

4. Why is PQ in the set R? 

5. Does this prove that R is a convex set? 

6. Does this prove the following theorem? 

THEOREM 2.1 The intersection of two convex sets is a convex 



Figure 2-26 suggests the first of our Separation Postulates. It shows 
point A dividing or separating line / into two convex sets T S and T, 



4 * • • • ► t 

B A C D 

Figure 3-36 

No point of / is in both S and T, Point A is the only point on line I that 
is in neither S nor T, If points B and C are in different sets (that la, if 
B £ § and C £ T, or if C £ S and R £ T) t then EC contains point A* 
On the other hand, if points C and D are in the same set (that is, if 
C £S and D £ S, or C £ T and D £ T ), then CD does not contain 
point A. 

We state these ideas formally in our next postulate. 

POSTULATE 12 (The Line Separation Postulate) Each point 
A on a line separates the line. The points of the line other than the 
point A form two distinct sets such that 

1. each of the two sets is convex; 

2. if two points are in the same set. 
then A is not between them; 

3. if two points are in different sets, 
then A is between them. 



2.5 The S*pa ration Postulates 59 



Definition 2.7 Let a line I and a point A on I be given, 

1. The two convex sets described in Postdate 12 are called 
hairline* or sides of point A on line I; A is the endpoint of 
each of them. 

2. If C and D arc two points in one of these sets, we say that 
C and D are on the same side of A, or that C is on the 
D-side of A, or that D is on the C-side of A. 

3. If B is a point in one of these sets and C is a point in the 
other set, we say that B and C are on opposite sides of A 
on line / or that B and G are in the opposite halftones of l 
determined by the point A. 



Note that a halfline is a ray with the endpoint omitted. Although it 
is convenient to think of a halfline as Juicing an endpoint, remember 
that a halfline does not contain its endpoint. (This use of "have" should 
not bother yon since you have friends, but you do not contain them.) 
Note also that opposite hairlines are collineur and that they have the 
same endpoint. 

Since there are infinitely many lines in space that contain any given 
point, it follows that although a point has only two sides on any given 
line, it has infinitely many sides in a plane or in space. Normally, we 
do not speak of the sides of a point except when a given line contains 
the point and we are discussing the sides of the point on that line. 

Postulate 13 describes how a line separates any plane containing 
die line. In Figure 2-27, we see that the points of plane « that do not 
lie on line / are separated into two sets: 

(1) those points that are on the B-side of line h 

(2) those points that are on the C-sidc of line /. 




« 



Figure 1-27 



60 Separation and Related Concepts Chapter 2 

No point is in both of these sets and no point of line I is in either of 
the two sets. If points B and Care in different sets, then 2?C intersects I. 
On the other hand, if points C and D are in one set, then CD does not 
intersect line L We state these ideas formally in our next postulate. 

POSTULATE 13 ( The Plane Separation Postulate) Each line I 
in a plane separates the plane. The points of the plane other than the 
points online / form two distinct sets such that 

1. each of the two sets is convex; 

2. if two points are in the same set, then 
no point of line I is between them; 

3. if two points are in different sets, then 

there is a point of line I between diem. 

Definition 2.8 Let a plane a and a line / in a be given. 

1. The two convex sets described in Postulate 13 are called 
halfplanes or sides of / in plane a; l is the edge of each of 
them. 

2. If C and D are two points in one of these sets, then we say 
that C and D arc on the *ame side of / in plane a t or that C 
is on the D side of t t or that D is on the C-stde of I or that C 
and D are in the same h airplane. 

3. If B is a point in one of these sets and C is a point in the 
other set, wc say that B and C are on opposite sides of / in 
plane a or that B and C are in the opposite halfplanes of a 
determined by the line L 

Note that a halfplane does not contain its edge and that opposite 
halfplanes are coplanar and have the same edge. Furthermore, since 
there are infinitely many planes in space that contain any given line, it 
follows thai although a line has only two sides in any given plane, it has 
infinitely many sides in space. Normally, we do not speak of the sides 
of a line except when some given plane contains the line and wc arc 
discussing the sides of the line in that plane. 

Figure 2-28 suggests the last of the Separation Postulates. The 
figure shows how a plane a separates the points of space that do not lie 
in plane a into two sets: 

(1) those points that arc on the B-side of plane a; 

(2) those points that are on the C-side of plane a. 

No point is in both of these sets and no point of plane a is in either 
of the two sets. If points B and C are in different sets, then BC inter- 



2,5 Th* Separation Postulates 61 




Figure £28 



sects plane a. If points C and D are in the same set, then CD does not 
intersect plane a. 

Although these space separation ideas can be proved using the 
postulates introduced up to this point we shall summarize and state 
these ideas formally as a postulate in order to simplify and shorten our 
formal development of geometry. Note the similarity of the Space 
Separation Postulate to the Plane Separation Postulate. 

POSTUIATE 14 (Tke Space Separation Postulate) Each plane 
tit in space separates space, The points in space other than the points in 
plane a form two distinct sets such that 

1. each of the two sets is convex; 

2. if two points arc in the same set, then 
no point of plane a is between them; 

3. if two points are in different sets, then 
there is a point of plane a between them. 






Definition 2.9 Let a plane a tie given. 

1. The two convex sets described in Postulate 14 are called 
half spaces or aides of plane a and plane a is called the face 
of each of them. 

2. If C and D are any two points in one of these sets, then we 
say that C and D arc on the same side of a., or that C is on 
the D-side of a, or that D is on the C-side of a, or that C 
and jD are in the same halfspaec, 

3. If B is a point in one of these sets and C is a point in the 
other set, then we say that B and C are on opposite sides 
of a or that B and C are in opposite halfspaces. 



62 Separation and Refated Concepts 



Chapter 2 



Note that a halfspace does not contain its face and that opposite 
halfspaces have the same face. Furthermore, although a point or a line 
has infinitely many sides in space, a given plane has only two sides in 
space. 

The statements of the Separation Postulates and the definitions 
accompanying them are lengthy. However, the ideas they convey arc 
simple and can be easily described by means of figures. Briefly, the 
Separation Postulates tell us that a point separates a line into two half- 
lines; that a line separates a plane into two halfplanes; that a plane 
separates space into two halfspaces; and that these sets are convex. 

Notation. Figure 2-29 shows point A on line I and two halftones de- 
termined by point A, Wc may denote the halHine on the C-side of A in 

c ► 

line / by the symbol U or by the symbol AC. Similarly, the halfline on 
the B-side of A may be denoted by I2 or by AB. Be careful to note the 

difference between the symbol AB for ray AB and the symbol AB for 
hairline AB. Does AB = a3? Explain. 



p 



C 



■*/ 



Figur«S- 28 



Figure 2-30 shows line I in plane a and the two halfplanes which 
line I determines. We may denote the halfplane on the /J-side of line 1 
in plane a by ffi, and the halfplane on the C-side of line I in plane a by 
a 2 . Similarly, if S represents the set of all points in space, we may de- 
note the two halfspaces into which space is separated by a plane with 
the symbols Si and S2. 



D« 




Figure 2-30 



2.5 The Separation Postulate* 63 

We make use of the Separation Postulates to prove the next two 
theorems, 

THEOREM 2,2 If m segment has only one endpoint on a given 
line, then the entire segment, except for that endpoint, lies in one 
half plane whose edge is the given line, 

RESTATEMENT: 

Given: Line I and segment AB in plane a such that / and AB have 
only the point A in common, 

To Prove: If X is any point of AB such that A-X-B, then X is on 
the B-sade of Mn plane a. 




Figure 2-31 



Proof. 



1, X is a point such that A-X-B. 

2, A, X, B are distinct points, 

3, I does not intersect XK 

4, X and B are not on opposite sides of I 

5, X and B are on the same side of /, 



1. Given 

2. Why? 

3. Why? 

4. Why? 

5. Plane Separation 
Postulate 



We have shown that X, which is any point of -AB except point A or 
point B, is on the B-side of t in plane a; hence every point of AB, except 
point A, is on the B-side of / in plane a. 

THEORFM 2.3 If the intersection of a line and a ray is the end- 
point of the ray. then the interior of the ray is contained in one half- 
plane whose edge is the given line, 

RESTATEMENT: 

Given: AB and line / i ntersect in j ust the point A in plane a. 
To Prove: All points of AB are on the B-side of / in plane a. 



64 Separation and Related Concepts Chapter 2 




Figure 2-38 

Proof: Let X be any point of AB such that A-B-X. 

1. All the points of All, except A, lie on the 2Mde of / in plane 
cl Why? 

2, X £ I Why? 

& I does not intersect BX. Why? 

4, X is not on the opposite side of I from B, Why? 

5, X is on the same side of t as B, Why? 

Since X is any point such that A-B-X, it follows that crop BA Ues 
entirely on the B-side of I Since AB is the union of opp BA and ?0f with 
A deleted, it follows that A ft is on the B-side of I 



EXERCISES 2.5 

1. Prove that the intersection of any three convex sets is a convex set, 

2. There is a theorem which "extends" Theorem 2.1 to any numl>er of sets. 
Stale this theorem. 

3. Is a ray a convex set? 

4. Is the interior of a ray ft convex set? 

5. Consider the following definition, of a ray: Ray AS consists of point A 

and all the points on the B-side of A on line AB. Define opp ~AB in a 
similar manner. 

6* Complete the following: Let SI be a segment. Then AB — AB f\ BA. 

Since AB and BA are convex sets it follows from [7] that [T] is a convex 

set. 

7. Let I be any line and let P and Q be any two distinct points on /. If X 
is any point between P and (>, is X on line I? Why? Does this prove that 
PQ is contained in I? Does this prove that every line Is a convex set? 

8* Let a be any plane and let Pund Q be any two distinct points in plane «. 

(a) Is PQ in plane a? Why? 

(b) Is PQ* subset of ?$* Why? 



2.5 The Sepa ration Postulates 65 

(c) Is PQ a subset of plane «? Why? 

(d) Does this prove that every plane is a convex set? 

9. Hie interior of each circle shown in the figure below is a convex set 
(a) Is the intersection of these interiors a convex set? Why? 

(h) Is l he union of these interiors a convex set? 




10, Draw two circles in such a way that the union of their interiors is a 
convex set, 

*— • * — » 

Exercises 1 1-15. In Figure 2-33, lines AB and CD intersect at E so that 

A-E-B and C-E-D. The D-side of a2 has been shaded. 




Figure £33 



11. Copy the figure and shade the B-side of 

12. Why are the two shaded half planes coplnnar? 

13. Why is the intersection of these two half planes a convex set? 

14. Describe in your own words the intersection of the two shaded half- 
planes. 

15. Does your description in Exercise 14 suggest a definition of "interior of 
an angle?" 

16. Let P, Q, and R be distinct points on a line L with ft and Q in the same 
hatfline with endpotnl P, On the basis of the given information is it pos- 
sible that P is between Q and R? That Q is between H and P? That R is 
between P and Q? 

17. Let F, Q, R be distinct points on a line /, with R and Q in opposite half- 
lines with endpoiut P. On the basis of the given information, is it pos- 
siWe that JP is between Q and R? That Q is between R and F? That R is 
between P and OP 



66 Separation and Related Concept! Chapter 2 

18. Is the union of two opposite halflmes a line? Explain. 

19. Is the union of two opposite halflines a convex set? 

20. Is the union of two opposite rays a line? Is their union a convex set? 

21. In what respect does the set of points in ray AB differ from the set of 

o— ► 

points in halflme AB? 

22. Is the interior of ray AB the same set of points as halfline a2? 

23. Is the u niou of two opposite h airplanes a plane? Ls their union a convex 
set? 

24. Line / lies in plane a, Point A is on one side of / in plane n. Point B is in 
plane a and is not on the A -side of I and is not on the opposite side of / 
from A. Make a deduction, 

25. A, B, C, X arc lour distinct points on line m, B and A are on opposite 
sides of A* and C is on the A -side of X. Draw a conclusion about points B 
and C. 

26. E, F y C are three distinct points in plane a. E and F are on opposite sides 
of line n in plane a. If E and C are on opposite sides of line ft, what con- 
clusion can you draw with regard to points G and F? 

27. From which postulate may we infer that a given line in a given plane 
Ixas only two sides? 

28. Explain why the following statement is true: If J* and Q are any two dis- 
tinct points in halfplane on, then FQ is in a*. 

29. A halfplane is an example of a "connected region," A line in a plane 
separates the points of the plane not on this line into two connected 
regions. Into how many distinct connected regions do two distinct inter* 
seeting lines separate the remaining points of the plane that contains 
them? 

30. Into bow many distinct connected regions do three distinct coplanar 
lines separate the remaining points of the plane that contains them if no 
point lies on all three lines and if each two of the lines intersect? 

Zh Into how many distinct connected regions do four distinct coplanar 
lines separate the remaining points of the plane that contains them if no 
three of the lines contain the same points and if each two of Lhe lines 
intersect? 

32. Use your answers to Exercises 29-31 to predict the number of distinct 
connected regions into which five distinct coplanar lines separate the 
remaining points of the piano that contains them if no three of the lines 
contain the same point and if each two of the lines intersect 

33. Can three distinct coplanar lines be situated so as to separate the re- 
maining points of the plane that contains them into three distinct con- 
nected regions? Four distinct connected regions? Five distinct con- 
nected regions? Six distinct connected regions? Seven distinct 
connected regions? More than seven distinct connected regions? 



2,6 Interiors and Exteriors of Angles €7 




Figure 2-.M 



34. Draw a figu re for each part of Exercise 33 to which yon answered "Yes. 

35. Figure 2-34 shows two distinct 

planes ft and /? intersecting in 
a line t The line Z is the edge of 
taw many different halfplanes 
represented in the figure? 

36. Name two distinct halfplanes 
represented in Figure 2-34 that 
are coplanar. 

37. Name two distinct halfplanes 
represented in Figure 2-34 that 
are not coplanar. 

38. How many different pairs of 
halfplanes in Figure 2-34 are 
not coplanar? 

39. Chic plane separates the rest of space into two connected regions. Into 
bow many distinct connected rejpons do two distinct intersecting planes 
separate the rest of space? 

40. Into how many distinct connected regions do three distinct intersecting 
planes separate the rest of space if no line lies in all three of the planes, 
if every two of the planes intersect, and if each plane intersects the line 
of intersection of the other two pla 

41. L'se your answers to Exercises 39 and 40 to predict the number of dis- 
tinct connected regions formed by four distinct intersecting planes if 
no three of these planes contain the same line, if each two of these 
planes intersect, and if each plane intersects each line of intersection 
formed by two of the other planes. 

42. challenge problem. Construct a model to represent the situation of 
Exercise 41. Count the number of distinct connected regions formed. 
How does this number compare with your prediction? 

43. challenge problem. Extend the result of Exercise 41 to five planes. 



2.6 INTERIORS AND EXTERIORS OF ANGLES 

In Section 2.5 we introduced the concept of separation. A line in a 
plane separates the points of the plane not on the line into two half- 
planes. A pie rure of an angle suggests that an angle separates its plane. 
Indeed, if plane a contains tLABC, then all the points of at that are not 
points of L A BC make up two sets, one called the interior and the other 
the exterior of Z ABC. We shall state carefully what we mean by these 
terms. 



Separation and Related Concepts 



Chapter 2 



One of the simplest ways ly think of the interior of an angle is as the 
intersection of two halfplanes associated with the angle, For /.ABC 






%ur*2-35 

these haifplanes are the C-sitle of AS and the Aside of BC, as indicated 
in Figure 2-35. Our formal definition is as follows: 



Definition 2.10 The interior of an angle, sa y £ABC t is 
the intersection of two haifplanes, the C-side of AB and the 
A-side of BC. 




Figure £38 



Definition 2.11 The exterior of an angle is the set of all 

points in the plane of the angle except those points on the 
sides of the angle and in its interior. 



Figure 2-36 illustrates both Definitions 2. 10 and 2.1 1. 
Tile following theorem is easy to prove using Definition 2.10 and 
Theorem 2.3, 



2.6 Interiors and Exteriors of Angles 69 



THEOREM 2.4 If P is any point in the interior of Z ABC, then 
the interior points of ray BP are points of the interior of Z ABC 

RESTATEMENT: 

Given: Z ABC with P a point in the interior of Z ABC. 
To Prove; B?is in the interior of /.ABC 



Proof: By definition of the interior of L ABC, 
P is on the C-side of AJB and on the A-side of 
BC. By Theorem 2.3, BP is on the C-side of AB 

* — » o • 

and on the A -side of BC. Therefore BP is in 

the intersection of these two halfplanes which, 
by definition, is the interior of /ABC. 




c 

Figure 2-37 



EXKRCISE5 2.6 

L Copy and complete the following definition. 

The interior of /PQR is the [7j of the haffplane lhatis the 
P-side of [f] and the halfplane lhat is the \?j oi PQ. 

Exercises 2-7 refer to Figure 2-38. 




2* Which of the labeled points are in the interior of Z ABC? 

& Which of the labeled points arc in the exterior of Z ABC? 

4. Which of the labeled points are not in the interior of L ABC? 

5. Which of the labeled points are in the interior of Z GBR? 

6. Which of the labeled points are in the interior of Z G7J/JP 

7. Which of the labeled points are in the interior of Z CBG? 



70 Separation and Related Concepts Chapttr 2 

§, Is the vertex of an angle u point of the interior of the angle? Explain. 
9. Is the vertex of an angle a point of the exterior of the angle? Explain, 

10. Is B a point of the interior of Z ABC? Explain. 

11. Is B a point of the exterior of LABC? Explain. 

12. Suppose that A, B, C, D are four noncoplanar points. Is it possible that 
D k a point of the interior of <LABC? Explain. 

13. Suppose that A, B t C, D are four noncoplanar points. Is it possible that 
D is a point of the exterior of £ABC? Espial 

■ Exercises 14-20 refer to Figure 2*39. 




Kipire 2-3S 

14. Make a copy of Figure 2-39 and shade the A^side of F.B with vertical 
raw ff and the £-sfde of AB with horizontal rays ^, 

15. Which angle has only vertical shading in its interior? 

16. Which angle has only horizontal shading in its interior? 

17. Which angle has both kinds of shading in its interior? 
IS, Which angle has no shading in its interior? 

19. Is the intersection of the I wo shaded portions the interior of any of the 
angles shown in the figure? If so, name the angle(s). 

20. Is the union of the two shaded portions the exterior of any of the angles 
shown in the figure? The interior? 

21. Does au angle separate the points of its plane not on the angle into two 
connected regions that do not intersect? 

22. What axe the connected regions of Exercise 21 called? 

23. Draw a picture of a n angle. Mark two points Pand Q hi the exterior of 
the angle. Does PQ intersect the angle? 

24. Does your answer in Exercise 23 depend on your choice of P and Q or 
would the answer he t he same for every choice of P and Q? 

25. Is the exterior of an angle a convex set? 

26. Draw a picture of an angle. Mark two points P and Q such thai P is 
in the interior of the angle and Q is in the exterior. Does PQ intersect 
the angle? 



2,6 Interiors met Exteriors of Angles 71 

27. Does your answer in Exercise 26 depend on your choice of F and Q t or 
would the answer be the same for every choice of P and Q if P is chosen 

in the interior of the angle and Q in the exterior? 

28. challenge PimiiLEM. It can he proved that if P is a point in the in- 
terior of an angle and Q is a point in the exterior,, then PQ intersects the 
angle. The proof is difficult and there are essentially five cases to con- 
sider as indicated in Figure 2-40 where £ ABC is the angle, P is a point 
in the interior of the angle, and Qi, Qs, Q& Q± t Q& are points in the ex- 
terior of the angle such that 

■i — * *— i ■» 

(1) Qi is any point on the A -side of BC and on the opposite side of AS 

from C. 



(2) Q 2 is any point on Vpp B( 

^ y 

(3) Q3 is any point on the opposite side of AB from C and on the oppo- 
site side of BC from A* 

{4} Qa is any point on &pp BA, 

(5) Qs, is any point on the C-side of AB and on the opposite side of BC 
from A. 



* 



/ 



■% 




Wjjure 240 



Following is an outline of a proof for case 1, a proof that if Qi is any 
point on the A-xidc of BC and on the opposite side of AB from C, then 
~F^fi intersects LABC. Answer the "Whys" in this outline, 
i. P and C are on the same side of A H. Why? 

2. C and Qi are on opposite sides of AB. Why? 

3. P and Qi are on opposite sides of AB* Why? 

i — » 

4. There is a point R between P and Qi on line BA, Why? 

<■ — * 

This means that PQi intersects AB in an interior point of PQi . We wish 

now to prove that PQ\ intersects BA rather than Qpp BA. 



72 Separation and Ralated Concepts Chapter 2 

5. P Is oil the A-side of M Why? 

6. Q t is on the A-side of K?. Why? 



7. The A-side of BC is a convex set Why? 

B. AH points of FQx are on the A-side of KJ. Why? 

Therefore the point fl is on the A-side of BC and on M. Therefore R 
is on BA, and hence H is a point of I ABC. Therefore PQi intersects 

labc. 

29. challenge pboblkm . Let the situation of Exercise 28 be given. Prove 
that PQ 2 intersectsTJA. 

30. CHALLENGE problem. Let the situation of Exercise 28 be given. Prove 
that P{) s intersects L ABC. 

31. Draw a picture of a n angle. Mark two points F and Q in the interior of 
the angle. Does PQ intersect the angle? 

32. Docs your answer in Exercise 31 depend on your particular choice of 
P and Q? 

33. Is the interior of an angle a convex set? 

34. challenge pkoblem. Use Theorem 2.1, the Plane Separation Postu- 
late, and the definition of interior of an angle to prove that the interior 
of an angle is a convex set. 

35. On the basis of your experiences in informal geometry try to write a 
definition of a triangle, thinking of it as a set of points. (Remember that 
a segment is a set of points and that the union of several sets of points 
is a set of points.) 

36. On the basis of your experiences in informal geometry try to write a 

definition of a quadrilateral. 

2.7 TRIANGLES AND QUADRILATERALS 

Next to segments and angles perhaps the simplest geometrical fig- 
ures are the polygons, and the simplest polygons are the triangles. You 
all know what a triangle looks like. It has three sides and three angles, 
A drawing of a triangle (Figure 2-11) shows its three sides, which are 
the segments AB, BC, and GA. 




2.7 Trianghw and Quadrilaterals 73 

On the other hand, the angles of a triangle are not shown completely 
in a picture of the triangle. In Figure 2-42, however, there are pictures 
showing the tingles of a triangle. A picture that does show all the angles 
of a triangle is not a picture of what is usually meant by a triangle. 




its uigto* 



Figure 2-42 



The following definition is worded carefully so that it says exactly 
what we want it to sav. 



Definition 2.12 If A, B f C are three noncollinear points, 
then the union of the segments A B, EC, CA is a triangle. 



Notation. The triangle which is the union of the segments AB, BC, 
CA is denoted by AABC 

x\n alternate definition which you might prefer is as follows. 

► A triangle fe the union of the three segments determined by 
three noncollinear points. 



This is an acceptable definition since we have agreed (in Section 
2.3) that "if A, B. C are three distinct points, then the segments de- 
termined are AB, BC, and CA." 



74 Separation and Related Concepts 



Chapter 2 



Definition 2J3 Let A AiJC be given. 

1. Each of the points A, B. C is a vertex of A ABC 

2. Each of the segments AB, W, CA is aside of A ABC. 

3. Each of the angles L ABQ Z BCA, L CAB is an an^e of 
AABC. 

4. A side and a vertex not on that side are opposite to each 
other. 

5. A side and an angle arc opposite to each other if that side 
and the vertex of that angle are opposite to each other. 



Note that a triangle contains its vertices and its sides but that it does 
not contain its angles. An angle of a triangle is not a subset of the tri- 
angle. It is customary to speak of the "angles of a triangle" or "the 
angles determined by a triangle," but it is incorrect to refer to them as 
"the angles contained in the triangle." Remember that whereas a tri- 
angle has angles, it does not contain them. (In this connection it might 
be helpful to think of a farmer who has farms and barns but does not 
contain them, or of a man who has a car and a house and lot but does 
not contain them!) 

Figure 2-43 illustrates the interior of a triangle and the interiors of 
the angles of a triangle* 





AA HC, l~A t interior of L A 



AABC, jLB, interior of Z_B 





AABC, LC, interior of £C 
Figure £-43 



AABC, interior AABC 



2.7 Triangle* and Quadrilateral* 75 



Definition 2.14 The intersection of the interiors of the 
three angles of a triangle is the interior of the triangle. 



Definition 2.15 The exterior of a triangle is the set of all 
points in the plane of the triangle that are neither points of 
the triangle nor points of the interior of the triangle. 



In Figure 2-44 the interiors of Z A and L B are shaded. Note that 
the intersection of these interiors is the interior of the triangle. 




figure 2-M 



This Suggests the following theorem, 

THEOREM 2. 5 The i ntersecrion of I he interiors of t wo angles of 

u triangle is the interior of the triangle. 



Proof: Let A ABC be given. I^et X be any point in the intersection of 
the interiors of any two of its angles, say L A and Z B. 

X is on the B-side of AC because it is in 

the interior of LA. 

X is on the A -side of BC because it is in 
the interior of Z B. 

X is in the interior of Z C because it is on 

die B-side of AC and the A-side of BC. 

This proves that if a point is in the interiors of two angles of a tri- 
angle, then it is also in the interior of the third angle. Therefore the in- 
tersection of the interiors of two angles of a triangle is contained in the 



76 Separation and Related Concepts Chapter 2 

interior of the triangle, Since the interior of the triangle is the inter- 
section of the interiors of all three angles, it follows that the interior of 
the triangle is contained in the intersection of the interiors of any two 
of its angles. Therefore the interior of the triangle is contained in, and 
also contains, the intersection of the interiors of any two of its angles. 
Therefore the interior of the triangle is the intersection of the interiors 
of any two of its angles. 

Consider any three n on colli near points A, B, C in a plane a and a 
line I in a which does not contain A or B or C but which does contain 
an interior point of AC as shown in Figure 2-4*5. 




Figure fc-45 

L B is either on the C-side of I or on the A-side of / in plane a. 
Why? 

2. If B is on the C-sidc of I, then I intersects AB in a point in the 
interior of AB, Why? Does I intersect BC in this case? Why? 

3. If B is on the A-side of /, then I intersects BT in a point in the 
interior of BC. Why? Does / intersect AB in this case? Why? 

4. Does this prove the following theorem? 

THEOREM 2,6 If a line and a triangle are coplanar, if the hue 
does not contain a vertex of the triangle, and if the line intersects 
one side of the triangle, then it also intersects just one of the other 
two sides. 

Triangles have three sides. Quadrilaterals have four sides. By quad- 
rilaterals we mean some., but not all, plane figures that are made up of 
four segments as suggested by Figure 2-46. 

Try writing a definition of quadrilateral before reading further. 
Then compare your definition with the following definition which we 
adopt for our formal geometry. 



2.7 Triangles and Quadrilaterals 77 




Quadrilateral* 





Not quadrilftt*rjiln 



Figure 2-46 



Definition 2. 16 Let A, B T C, D be four coplanar points such 
that no three of them are collinear and such that none of the 
segments AB, ffll, CD, DA intersects any other at a point 
which is not one of its endpoints. Then the union of the four 
segments AB, B£, UD, DA is a quadrilateral. Each of the four 
se^nents is a side of the quadrilateral and each of the points 
A, B, C, D is a vertex of the quadrilateral. 






Is it possible that four given points are the vertices of more than 
one quadrilateral? Figure 2-47 shows that this is indeed possible. Think 
of the figure as four different pictures of the same four points. The sec- 
ond, third, and fourth pictures show different quadrilaterals with the 
same vertices. 



p 



A 



E 






ASDC 



a urn 



ADBC 
Figure 2-47 



To name a quadrilateral using the names of its vertices, and to do 
it so that we know which segments are its sides, the names of the ver- 
tices are so arranged that { 1) letters adjacent to each other in the name 



78 Separation and Related Concepts Chapter 2 

of the quadrilateral are names of the endpoints of a side of the quadri- 
lateral and (2) the first and last letters in the name of the quadrilateral 
arc names of the endpoints of a side of the quadrilateral. 

Note that if AB CD is a quadrilateral, then BCDA is the same quad- 
rilateral. Give several additional names for this quadrilateral. Notice 
also that ABCD and ACBD are not names for the same quadrilateral. 



EXERCISES 2.7 

1. Copy and complete the definition: If A, B, C are three [?] points, then 
is the triangle denoted by A ABC. 

2. Name the side which is opposite lo /. ABC in A ABC. 
% Name the angle which is opposite to side AB in A AUG'. 

In Exercises 4-10, let A ABC be given. State whether or not the given set 

is a subset of A ABC, 

4. AABC 8. AB U BC J CA 

5. Interior af / ABC 9. AB f BC ft 25 

6. Interior of AABC 10. {A, B s C} 

7. AB 

In Exercises 11-1-5, let AABC be given. Let S denote the union of AABC 
and its interior. State whether or not the given set is a subset of S. 

11. Interior of £ABC 

12. (Interior_of_/ ABQ D (Interior of IBCA) 

13. [AB.BC, CA) 

14. AB HBCnCA 

15. AB n (Interior of A ABC) 

In Exercises 16-19, let AABC be given. Let II U H 2x H 3 denote the follow- 
ing halfplancs, respectively: the A -side of BC, the B-side of CA, and the 
C-side of AB. Express the given set in terms of these halfplanes. 

16. Interior of £ABC 18. Interior of ZCAB 

17. Interior of LBCA 19, Interior of AABC 

In Exercises 20-24, let the same situation as in Exercises 16-19 be given. 

State whether or not the given set is a subset of the exterior of AABC. 

20. Tlie interior of the ray 22. BC 
opposite to BC 23, BC 

21. The ray opposite to BC 24. The opposite halfplane front //j 



2.7 Triangles and Quadrilaterals 79 

25. Draw a picture of a triangle and a line so lhat 

(a) their intersection is one point; 

(b) their intersection is exactly two points; 

(e) their intersection consists of more than two points. 

26. Gun a line intersect a triangle in exactly one point which is not a vertex? 
Illustrate. 

27. What is your answer to lixercise 26 if the line and triangle arc contained 
in the same plane? 

28. Can a line intersect a triangle in exactly three points? 

29. Name all of the triangles shown in the figure. 

D 




30. If A , B t C are the vertices of a triangle, prove that there is a point of L A 
that is not a point of the triangle. 

31. If A, H f C are noncollincar points, is the following statement true? 

AABC = {I A n IB) U {I A n Zqj(ZBn ZC> 

32. Let A ABC with points P and {> such that A-F-B and A -Q-C as in the 
figure be given. To prove that A ABC is not a convex set of points, it is 
sufficient to show that there is a point X of fQ which is not contained 
in A ABC. Let X be a point of TQ such that P-JE-Q, 

A 




(a) X £ AC. 

(b) X € AB. 

(c) X £ BC. Why? {Hint: Use Tlicorem 2.6.) 
Does this prove that B triangle is not B convex set? 

33. Is the exterior of a triangle a convex set? Illustrate with a figure. 

34, Is the interior of a hi angle a convex set? Kxplain why. 



80 Separation and Related Concepts Chapter 2 

■ For Exercises 35-38 draw a triangle, A ABC. 

35. Mark a point that is in the interior of Z A and in the exterior of the 
triangle. 

30* Mark a point that is in the interior of Z A but is neither In the interior 

nor the exterior of the triangle. 

37. Mark a point in the exterior of the triangle that is not in the interior of 
any of the angles of the triangle. 

38, Can you mark a point that is in the interior of Z A and is in the interior 
of ZB but is not in the interior of ZC? Why? 

■ In Exercises 39-42, let quadrilateral ABCD be given. 

39, Is it possible that D is an element of BC? 

40, Is it possible that D is an element of the interior of L BAC? 

41. Is it possible that D is an element of the exterior of Z BAC? 

42. Is it possible that the interior of A ABC is a subset of the interior of 
A A DC? 

43. Prove that every plane contains a quadrilateral, 

44, Prove that any three vertices of a quadrilateral are the vertices of a 
triangle. 



2.8 PROPERTIES OF EQUALITY AND NUMBER OPERATIONS 

In preparation f or Chapter 3, which involves some algebra, we con- 
chide Chapter 2 with several sections devoted to a review of elemen- 
tary algebra. Chapter 3 contains many equations, that is, statements 
of equality. Following arc some basic properties of equality and num- 
ber operations that are useful in working with equations involving 
numbers. 

Substitution Property of Equality. In any statement about some 
thing (physical object, number, point, line, idea, etc) one of the 
names for that thing may be replaced by another name for the same 
thing. If the original statement is true, then the statement with the 
substitution made is also true. If the original statement is false, then 
the statement with the substitution made is false. 

Recall that A = B means that A and B are names for the same 
thing. The substitution property tells us that we may replace A by B in 
any statement about A without changing the truth or falsity of the 
statement. 



2,8 Equality and Number Operations 81 
Example Consider these two statements: 

(1) 7 = 2 + 5 

(2) 3 + 4 = 7 

From (1) and (2) it follows by the substitution property of equality 

3+4 = 2 + 5. 

Reflexive Property of Equality. For any thing a, a = n. 

Symmetric Properly of Equality. For any thing a and for any thing 
b t if a = h, then h = a. 

Transitive Property of Equality. For any things a, b, c, if a = b and 
b = c, then a = c 

Addition Property of Equality for Numbers. If a, h> c are real num- 
bers, and if a = h y then 

a + c = b + c. 

More generally, if a, fc T c, d are numbers such that a = b and c = d* 
then 

a + c = fr + d and a — c= b — d. 

(It is appropriate to include subtraction here since each difference 
can be expressed as an addition. Thus a — c — a + (— c). Also if 
c = d, then — c = —d.) 

Multiplication Property of Equality for Numbers. If a, b, c arc 
real numbers and if a = b t then 

ac = be. 

More generally, if a, b> c t d arc numbers, and if a = /; and c = d, 
then 







ac 


= bd t 


and if i 


c=£0,d=£Q f then 










c 





(It is appropriate to include division here since each quotient can 

be expressed as a product . Thus — = a* — - Also, if c = d* c =£■ 0, 

c c 

<f=^0\ then 1 = 1) 
c d 



82 Separation and Related Concepts Chapter 2 

Commutative Property of Addition. If a and b are real numbers, 
then 

a -f b m b 4* a. 

Commutative Property of Multiplication. Tf a and h arc real num- 
bers, then 

ab = ha. 

Associative Property of Addition. If a, h t c are real numbers, then 

(a + h) + c = a + {h + c). 

Associative Property of Multiplication, If a z /?, c are real numbers, 
then 

[ab)0 = a(bc). 

Distributive Property of Multiplication over Addition, If a, h, c are 
real numbers, then 

a(b + c) = ab + ac and [a + b)c = ac 4- fee, 

EXERCTSES 2.8 

■ In Exercises l-20 t name the property that justifies the given statement 

L tf x = 5 and x = y. then, y = 5. 

2, If ac = 5 and 5 = y, then x = y, 

a 7 = 7 

4. If AB = CD, then CD = AB. 

5. If 3 = x and 4 = y, then 3 + 4 = x 4- y. 
6* if 3 = x and 4 = y, then 12 = xy, 

7. If § is a coordinate system, then $ = g. 

8. If x 4- y = z and if x = a + h, then (a + b) 4- y a ft 

9. If lit = tcandif a 4- fr = uandc 4- d 1 = c, then (a 4- fr)(c -f tt) = 10. 

10. If -£ = c, tiien a = c/j. 
o 

IL Ifa + fc=c, then (o 4- fc) 4- (-&) = c + (-b). 

12. If f";? _ 6 lhcn x _ 5 _ ^ _ Q j 

13. 3(20 + 5} = 3 • 20 4- 3 ■ 5 

14. 3(xy) = (3% 

15. (2 4- 3)5 = 2 • 5 + 3 • 5 

16. (2 4- 3){4 4- 7) = (2 4- 3)4 4- (2 4- 3)7 

17. (3 + 2-3} + 7 = 3 4- (2-3 4- 7) 



2.9 Sohrfmg Equation* 83 

IS. 5 + i=i + 5 

19. 3 + (4 + 7) = 3 + (7 + 4) 

20. 3{7 + x) = (7 + x)3 



2.9 SOLVING EQUATIONS 

The properties of equality and number operations are useful in solv- 
ing elementary equations. 

Example 1 Solve for xt x + 3 = 52. 

Solution: 

x + 3 = 52 Given 

x = 49 Addition property (Acid -3) 

Example 2 Solve for *; 3,t - 5 = 7x + 2. 

So?u£ion: 



&r - 5 = 7x + 2 


Given 


3:t = 7* + 7 


Addition property (Add 5) 


-4x= 7 


Addition property (Add — 7x) 


i= -I 
4 


Multiplication properly (Mult 
by -i) 


Example 3 Solve for 


x - 5 x + 8 



10 12 

Solutions 



r — a _ a: + 
10 12 



Given 



12(x - 5) = J0{x + 8) Multiplication properly (Mull. 

by 120) 
12r — 60 = 10* + 80 Distributive property 

12* = lOt + 140 Addition property (Add 60) 

2* - 140 Addition propei+y (Add — lOx) 

* = 70 Multiplication property ('Mult. 

by^> 



84 Separation and Related Concepts 



Chapter 2 



Example 4 Solve for x\ 



X — o 

x-7 



(**T) 



Solution: 

X-5 _ 5 
*- 7 " 8 

8(x - 5) = 5(x - 7) 

fix - 40 = 5* - 35 
3* =5 



X = 



5 



Given 

Multiplication property (Mult 

by8(*-7)) 

Distributive property 

Addition property (Add 
( - 5x + 40)) 

Multiplication property 



(Mult, 



JErompte 5 Solve for « x + ° = * * + 5x. 

^ o 



Solution; 

Jl±A = JLzi1+5x 
2 3 

3(* + 5) = 2(x - 1) + 30x 



+ 15 = 2x - 2 + 30x 
-29*= -17 



x = 



XL 

29 



Given 

Multiplication property (Mult, 

by 6} 

Distributive property 

Addition property (Add 
(-32a: - 15)) 

Multiplication property (Mult. 

by -i) 



Although the list of properties does not explicitly include the divi- 
sion and subtraction properties of equably for numbers, they are in- 
cluded implicitly. For example, dividing by 7 is the same as multiplying 
by \, and subtracting 5 is the same as adding —5, In Example 1 you 
could think of subtracting 3 and justifying it by the subtraction prop- 
erty instead of adding — 3 and justifying it by the addition property. 
In Example 2 you could obtain the last step from the preceding step by 
dividing by —4 and justifying it by the division property instead of 
multiplying by — j and justifying it with the multiplication property, 
Note, however, that the multiplication property does not permit us 
to divide by 0, since is not the reciprocal of any number. 



2.9 Solving Equations 
Example 6 Solve for x: a(x + x t ) = 5, where a =£ 0. 



Solution: 

a(x + «J = 5 

* + x x - ^ 
a 



x = — _ Xl 



Given 

Division property (Divide by a) 

Subtraction property (Subtract 



EXERCISES 2.9 



In Exercises 1-20, solve for x. Name the properties that justify the steps in 
your solution. Express your answer in simplest form. 



1. 3x + 3 = 2x + 4 

2. 3s* - 4 - 4x + 3 

3. i* = J*+i 

4. 0,75* - 100 

5. 4(x - 64) = 28 

i±4£«a 

5 

- 3(*~1) _. 1 

' s s 

" 8-5 -9 -(-6) 

ft x ~ l - 'J ~ 3 

* 7 - I " 15 - 3 

a: - 10 fc - () 



11, 4x - 5 = 14c - 75 

_1_3 - _]0-(-8j 

ia x 4- 4 = 40 - * 

*-7 „ 7 -(-3) 




14. 



10. 



-11 - 10 



5- 1 1 

15. 3{x + 15) = 2{x + 15) 

16. 3(.v + 15) + 2{ X + 16) 

17 '7^ocH^ 100 > 

ia ~5— = 7 + '^0^ 
20. 0.3x 4- 0.8x a 220 



1 - 



In Exercises 21 -40, solve for x. Express answers in simple form. 



21. 


3* + 


5 BS 






22. 


3(x- 


1) + 2(* - 


-1) 


= 4* 


23, 


5* + 


7 = 3{* + 


4)- 


5 


24. 


x — 
2 


1 _ x-3 
3 






25. 


2ar- 


3 5x- 


1 





26. 



*- 1 



|-£*** 



27. ox -1=4 



2S. 



g- 1 
1 

x- 1 



I* ~ ? 
5a:— 1 
7-3 

x-2 
3 



86 Separation and Related Concept* 

30. 5x ~ 1 = ^nl 
6* - 7 6x - 8 

31. * + % =3 

32. 3x + y = 

34. x-3 = Jk + 4 

35. 5( j - 3) = 3{* + 4) 



Chapter 2 



36. 5* - 3 = 3* + 4 



37,*~ 



k + 2 



2 3 

38. 4x - 5 = 3«i + 4xa 
x — *i 5 



39. 



*2- ari 
*+! = 



7 
-5 



2.10 EQUIVALENT EQUATIONS 

In solving an equation, wc find the number (or numbers) that 
"satisfies" the equation. Such a number is a root of the equation. In 
Example 1 of Section 2.9 the root of the equation 

x + 3 = 52 

is 49. Note that 49 satisfies the equation since 49 + 3 as 52 is a true 
sentence and that no other number satisfies the equation. The set of 
all roots of the equation x 4- 3 = 52 is 



since 49 is the one and only root of the equation, {49} is the solution 
set of the equation x 4- 3 = 52. 

In each example of Section 2.9 we used properties of equality and 
number operations to obtain other equations that have the same solu- 
tion set Equations that have the same solution set are called equivalent 
equations. Note in Example 2, for instance, that 



anc 



3x 


-5 = 


7* + 2 




3ac = 


7x + 7 




-4x - 


7, 




x — 


7 
4 



are four equivalent equations. The solution set of each of them is 

HI- 

Sometimes the properties of equality and number operations are 
used to produce equations not equivalent to the given equation. 



2.10 Equivalent Equations 87 
Example I 

x = 5 Given 

0=0 Multiplication property (Mult, by 0) 

It is true that 

if x = 5, then = 0. 

Indeed = is a true sentence regardless of what may be true about x. 
The solution set of x = 5 is obviously {5}. The solution set of = 
is the set of all real numbers. Even though it is true that 

if x = 5, then = 0, 
it is not true that 

x = 5 and = are equivalent equations. 

When we went from x = 5 to = 0, we went from an equation with 
one root to an equation with infinitely many roots. We did not lose any 
roots, but we certainly gained many of them I 

Example 2 

■■«=T +1 

x(x- 2J= {*- 2) + l(x-2) 
x* - 2* = x- 2 + x- 2 
x 2 - Ix = 2x - 4 
x 2 - 4x + 4 = 
(x - 2)2 = 
x~2 = 
x = 2 

In this example we gained a root somewhere along the way. Trie 
solution set of x = 2 is obviously {2}> but 2 is not a root of the original 
equation. Why? The solution set of the given equation is the null set. 

If we try to reverse the steps in this "solution/' we can Justify each step 
except the last one. Given x = 2. we cannot justify going from 

x{x - 2) = (x - 2) + l(x - 2) 

to * = 1- -f 1 by dividing by x — 2, The multiplication property 

permits us to divide by any number except and x — 2 = if x = 2. 



88 Separation and Related Concepts Chapter 2 

Example 3 

x 2 = 3x 
* = 3 

In this example we multiplied both sides of x 2 = 3* by — (or divided 

both sides by x). In doing this we lost a root. How did this happen? Di- 
viding by x is legal except if x = 0. Is a root of the given equation? 
Is 2 = 3 • 0? Yes, it is. The solution set of the given equation Is {0, 3}. 
The solution set of x = 3 is {3}. The equations x 2 = 3* and x = 3 
are not equivalent equations. 

Tn solving BQ equation through a seqin.-:iu.-i' <j| equations it is advis- 
able to check each step for a possible loss of roots. If there is a value of 
x that might be lost as a root (as is lost in going from x 2 = 3x to x = 3 
in Example 3), it should be identified and checked by substitution in 
the original equation. Of course, it is a root if and only if it satisfies the 
original equation. 

To make sure that no roots are gained in the solution process you 
may (I) check each root of the final equation by substitution in the orig- 
inal equation or (2) check to see if the equations can be obtained in re- 
verse order without a loss of roots at any step. If there is no loss going 
backward, there is no gain going forward. If a root is lost going back- 
ward, it was gained going forward, and hence is not a root of the given 
equation. In Example 2 we have a sequence of equations and there is 
no loss of roots in going forward. When we reverse the steps, there is 
no loss of roots at any step until the last one. 2 is a root of 

x(x - 2) = x - 2 + l(x - 2) 

x — 2 
but not a root of x = + 1< 

x — 2 

2 is lost as a root going backward; it was gained as a root going forward. 
Example 4 

1 = I Given 



X — i 

x — 1 = x - 7 Multiply by x — 7 

0=0 Subtract x — 7 

In this example we have a sequence of three equations and the so- 
lution set of the last one is the set of all real numbers. In reversing the 



2.10 Equivalent Equations 89 
steps we do not lose any roots until the last step. To get 

x — 7 

from x — 7 = x — 7 we divide both sides by x — 7, and this is not 
permissible if x s= 7. Checking, we see that 7 is not a root of the given 
equation. The solution set of the given equation is the set of all real 
numbers except 7. 

Example 5 n\ y — 2 _ _ 

x — 2 

(2) y - 2 = 7(* - 2) 

In this example there are two variables, x and y. In Equation (2) 
we may wish to think of x as the independent variable and y as the de- 
pendent variable. A solution f Equation (2) is an ordered pair of num- 
bers {a, b) such that Equation (2) is satisfied when x is replaced by a 
and y is replaced by b. Thus (3, 9) is a solution of Equation (2) since 

9 - 2 = 7(3 - 2) 

is a true sentence. 

If we solve Equation (2) for gf, we get 

y = 2 + 7(* - 2) or y = 7x - 12. 

Let a be any real number whatsoever. Then x = a and y = la — 12 
satisfy Equation (2). For upon substituting a for x and la - 12 for y 
in (2) we get 

(la - 12) - 2 = 7(0 - 2), 

which is a true sentence. Indeed, the set of all ordered pairs 
(a, la — 12), where a is a real number, is the solution set of Equation 
(2). Any other letter can be used for the symbol a here. Thus we could 
say, if we wish, that the solution set of (2) is the set of all ordered pairs 
(x> 7x — 12), where % is real. 

Is the set of all ordered pairs fa la - 12) also the solution set of 
Equation (1)? Let us check Substituting a for x and la - 12 for y in 
Equation (1), we get 

(7a - 12) - 2 - 
a -2 

or after simplifying, ' ( fl ~ ) = 1. This is a true statement for every a 

a — 2 



90 Separation and Related Concepts Chapter 2 

with one exception. It is not true if a = 2, The solution set of (1) is the 
set of all ordered pairs (a, la — 12) for a ^ 2, Equations (1) and (2) 
are not equivalent 

Some of you have graphed equations like (2). You know that its 
graph is a line. What is the graph of (1)? The graph of (1) includes all 
the points of the graph of (2) except the point with abscissa 2, that is, 
die point (2, 2). The graph of (!) is the union of two opposite halflincs. 
Or to use a bit of informal language, a line with a hole in it. 



EXERCISES 2.10 

In Exercises 1-20, check by substitution to see if the given value of x is 
root of the given equation, 

1. X 2 + 5x + 6.25 = 0, x = 1 

2. a* + 5x + 6.25 = t x= -2 

3. x* 4- 5* 4- 6.25 = 0, t - -2.5 

4. 3(x - 5) + 7(* - 5) = 10(x - 5), x = 5 

5. 3(x - 5) + 7(x - 5) = 10(x- 5), x = 5.1 

6. 3(x - 5) + 7(x- 5) = 10(x -5),i= 137.3 

7 x - 2 _ 2x + 3 T _ , 
" 6x~T9* 

= 2xJ4 2 



St- 


■ 6 


ar — 


2 


3x- 


-6 


x — 


2 


3x- 


-6 


x — 


2 


3x- 


-6 


1 





Gx + 9 

2*4-3 

fix - 9 



x= -1.5 



10. -*-=-% = -^4- * = 1°°° 
Gx + 9 

]£ _J_ = _^-,*=2 

x 4- 2 a : - 2 

x ■+■ 1 

14. 1 + 

x + 1 

1 



16. 



x + 1 

214-3 

x - 3 



x — 


1 


x a - 


l 


1 




2x 




X — 


1 


x a - 


l 


1 




2x 

X*- 




X — 


1 


1 


7- 


- 5 


x = - 


-6 



4-— l — = -^— t x= -1 



2.10 Equivalent Equation* 91 



x — 2 * + 3 i-4 

19.^+-*i| + -*^i=3.x=4 
x-2 T x + 3 x- 4 

2a x^ + JL ^3 x-4 = _ 3 

x — 2 x + 3 x— 4 



In Exercises 21-30, check to see if the two given equations are equivalent. 

If they are not equivalent, state whether there is a loss or gain of roots in 
going from the first equation to the second one. 

21. x* = -2x, x = -2 



22. 


2x=3,x 


= 1.5 








23. 


x = 5, x* 


= 25 








24. 


x = 5, x2 


= Sx 








25, 


x = 5, x - 


- 1 =4 








26. 


x = 5,2x 


■ I = 


11 






27, 


3x + 2x = 


: 6x, X 


= 






28. 


x-2 
3 


x- 3 
4 


4(x 


-2) 




29. 


x-2 

3 


x - 3 


x - 
x — 


-2 
3 " 


3 




4 J 


1 


30 


x-2 


x-3 


X — 


3 


4 




3 


4 ' 


X — 


2 " 


3 



3(x - 3) 



In Exercises 31-35, find an ordered pair of real numbers that satisfies 
second of the given equations but not the first. 



2-2 



X 



s 5, tj — S s 5x 



33. l^i = -^i, 2x - ij - 4 = 
x — 4 9 — 4 y 

^ x-2 = 2 x-2 =JL -2 



92 Separation and Related Concepts C-iarl-r „' 

■ In Exercises 36-45, find the solution set of the given equation. 

36. 3x - 4* = 8 - 5 

37. % - 2) + 7{x - 3) = 30 



42, M — £ a 2 
x— 1 



38. x + - + 1 = — 

x x 

39. ^ = x 
x 

40. x - 1 = x - 2 

41. 2* - 2 = 2(x - 1) 



43. * ~ * + «- 



45 



+ ^4=3 



x-1 x-2 x-3 
44. *~? = 5 



x-2 x-2 =2 / x-2 \ 
* x-3 x-3 \x-3^ 



CHAPTER SUMMARY 

There were three BETWKENNESS POSTULATES and three SEP- 
A RATI ON POSTULATES in this chapter. We list them below by name 
only. Try to state each of them in your own words. Draw a picture illustrat- 
ing what each postulate says. 

9, THE A-B-C BETWEENNESS POSTULATE, 

10. THE THREE-POINT BETWEENNESS POSTULATE. 

11. THE LINE-BUILDING POSTULATE, 

12. THE I-TXE SEPARATION POSTULATE. 

13. THE PLANE SEPARATION POSTULATE. 

14. THE SPACE SEPARATION POSTULATE. 

The following concepts were defined in this chapter. Be sure that you 
know all of them. 



SEGMENT 

INTERIOR OF A SEGMENT 

HAY 

INTERIOR OF A RAY 

OPPOSITE RATS 

ANCLE 

INTERIOR OF AN ANGLE 

EXTERIOR OF AN ANGLE 

TRIANGLE 

INTERIOR OF A TRIANGLE 

EXTERIOR OF A TRIANGLE 

QUADRILATERAL 



TWO SIDES OF A POINT 

ON A LINE 
TWO SIDES OF A LINE 

IN A PLANE 
TWO SIDES OF A PLANE 

IN SPACE 
HALFLINE 

OPPOSITE HALFUNES 
HALFPLANE 
OPPOSITE 1IALFPLANES 
ILYLFSPACK 
OPPOSITE H ALFSPACES 



Review Exerciset 93 

A set of points is called CONVEX if for every two points P and Q in 
the set, the entire segment ¥Q is in the set. The null set and every set of 
points that contains just one point are also said to be convex. Each of the 
following sets is a convex set: segment, ray, line, plane, h&lflinc, halfplane, 
balfspaee, interior of an angle, and interior of a triangle. 

Six theorems were stated and proved in this chapter. Study them again 
so that you know and understand what they mean. 

The last part of this chapter contains a review of elementary algebra. 
You should know and be able to use the properties of equality and number 
operations that arc useful in solving equations. 



REVIEW EXERCISES 

In Exercises 1-15, indicate whether the statement Is true or false. 

L If points A, B, C and line t arc in the same plane, and if A and B are on 
the opposite sides of J, then C" must be either on the A -side of I or on the 
B-side of I 

2. If B and C are two distinct points on the same side of line n in plane a, 
then every point of EC is on the B-side of n. 

3. UA-B-C (point B is between points A and C) and B-D-Q then A-D-C 
and A-B-D. 

4. If R-S-T and Q-S-T, then R-Q-S and R-Q-T. 

5. If points R and S are on opposite sides of line m in plane a and points 
R and T are on opposite sides of m, then S and T are on the same side 
of m. 

6. The betweenness relations R-S-T and R-U-T uniquely determine the 
order of the points R, S, T t U on a line, 

7. If two rays intersect, they have one and only one point in common. 

8. AJ9 = M. 

9.AB = bX 
10. opp A~S = BA 
1L AB = BA 

12. ab n bX = AB 

13. AB U B~X = A~B 

14. opp AB D opp B~A = A 9 

15. If point B is between points A and C, then BA and BC are opposite rays. 



94 Separation and Related Concepts Chapter 2 

■ Exercises 16-20 refer to Figure 2-4S. 

A 




16. How many angles are determined by the segments shown In the figure? 

17. How many triangles are determined by the segments shown in the fig- 
ure? Name them. 

18. Point E is in the interior of two angles. Name them. 

19. Only one of the labeled points in the figure is in the interior of each of 
three different angles. Name the point and name the three angles. 

20. Two of Ae labeled points in the figure are not in tlie interior of any of 
the angles. Name them. 

21. Draw A ABC. Mark a point D such that D is between A and B. Mark 
a point E such that B is between C and E. 

(a) Docs ED intersect AS? Why? 

(b) Does EZ> intersect BC? Why? 

(c) Does ED intersect A£? Why? 

22. Define the interior of £PQR> 

23. Cicen: Line 1 f£ line m. 

Point C is between points A and B oa liue L 
Point C is between paints D and E on line m. 
Point C is between points C and E. 
Point F is between points A and E. 

(a) Which point (C, G, or F } is In the interior of I ACE? 

(b) Indicate whether each of the following is true or false. 

(1) Point F is on line I, 

(2) Point F is on line iiu 

(3) Point F is a point of / ACE. 

(4) Point C is a point of L BCF.. 

(5) Line FG does not intersect AC. 



Review Exercise* 95 

24. If r f $, t, u are four distinct coplanar rays having a common endpoint 

and if no two of the rays arc collinear, how many angles are formed by 
these rays? 

25. Which of our postulates guarantees that a halfplane is a convex set? 

26. Is the intersection of two convex sets always a convex set? 

27. Explain why the interior of an angle is a convex set. 

28. Is a line with one point deleted a convex set? 

29. Is the. union of two convex sets always a convex set? 

30. Describe two convex sets whose union is a convex set. 

In Exercises 31-35, solve for x. 

31. 3x - 5 = Ix - 25 

32. 2(r - 5} ■ 3x - 5 

33.^1+1=^ 

34, 1.75x = 17.50 

35. x - 1 + 2(x - 1) = 17{* - 1) + 14 



In Exercises 36 --40, find the solution 


set 


36, 


x + x = 


2* 




37. 


x + 1 = 


x + 2 




38. 


X 






39. 


x-1 
2*-2 


_ 1 
" 2 




40, 


X - 1 = 


Iff* -9 







Chapter 




HeUa Hammld/Bopho GuiUumetttt 



Distance 

and Coordinate 

Systems 



3.1 INTRODUCTION 

At this point in our formal geometry we have do postulates con- 
cerning the sizes of objects. We have no basis for saying how big an 
object is or even for saying that one object is bigger than another. Al- 
though the word "size" is often used in informal speech* it is not in the 
official vocabulary of formal geometry. Instead we talk about the 
length of a segment, the measure of an angle, the area of a rectangle, 
the volume of a sphere, the distance between two points, and so on. 

In elementary mathematics it is customary to draw a horizontal 
number line with numbers increasing from left to right and a vertical 
number line with numbers increasing in the upward direction. In some 
applications it may be more convenient to order the numbers from 
right to left or from up to down. A number line, such as the one in Fig- 
ure 3-1 in which the numbers increase from left to right, is a good de- 
vice for illustrating certain relationships among real numbers. One of 
them is the order relation. The fact that 3 is greater than 1 is consistent 
With the fact that 3 is to the right of 1 in Figure 3-1. The fact that 1 is to 
the right of —2 agrees with the fact that 1 is greater than —2. 



! ! ► 

-4-3-2-1 I 2 3 4 5 6 Figure 3-1 



93 Distance and Coordinate Systems Chapter 3 

Another relationship is based on the betweenness relation for real 

numbers. Since — is between — ■ and %, the point marked — lies 
3 5 § 3 

between the points marked — and — on the number line shown in Fig* 
ure 3-2. 



Figure 3-2 

This chapter is about distance and coordinate systems. We begin by 
considering the distance between two points. The idea of a coordinate 
system on a line is an extension of our ideas about a number line. Co- 
ordinate systems are useful in developing the properties of distance. 

3.2 DISTANCE 

Asking "How long is a certain segment?" is equivalent to asking, 
"How far apart are the endpoints of that segment?" In the world of 
real objects we can answer the question, "TTow far apart?" by using a 
physical ruler. We might determine that two points P and Q are 2 yd. 
apart, or 6 ft. apart, or 72 in. apart 

To make a physical ruler graduated, say, in inches, we must know 
what 1 im is. We must have a segment 1 in. long. In the United Stales 
the accepted relation between inches and meters is 39.37 in. = 
1 meter. For many years the meter was described officially by two 
marks on a phttinum-iridium bar kept in France. These marks repre- 
sented the eildpointS of the segment that was the official meter. Al- 
though the modern standard for measuring distances is now based on 
the wavelength of a certain kind of light, the idea that a given segment 
may be a standard or unit for measuring distance is important in our 
geometry. Tn the remainder of this section we discuss informally some 
of the basic properties of distance and then state these properties for- 
mally as postulates. 

Given any segment, say FQ y we could agree that this segment is the 
unit of distance. Thus we might start with a given stick and say, "Let 
the distance from one end of this stick to the other end be 1, and let 
us call this unit of distance the "stick/' Then the given stick or unit is 
the basis of a system of distances which we might call the stick-system. 
In this system the distance between any two different points is a posi- 
tive number. In particular, the distance between the endpoints of the 



3.2 Distance 99 

unit stick is 1, Tf the distance between points R and S is 3 in this system, 
this means that R and S are 3 times as far apart as the endpoints of the 
unit stick. If the distance between points U and Vis 1 in this system, 
this means that V and Vare just as far apart as the endpoints of the unit 
slick. 

Speaking more formally, each segment PQ determines a distance 
function. The domain of this function is the set of all segments, or if 
you prefer, the set of all pah's of distinct points. The range of this func- 
tion is a set of positive numbers. It is convenient in developing the con- 
cept of distance to take the distance between a point and itself to be 0. 
Then the domain of a distance function is the set of all pairs (not neces- 
sarily distinct) of points and the range is a set of noimcgative numbers 
including 0. Later, after we adopt the Ruler Postulate, it will be obvious 
that the range is the set of all nonnegative numbers including 0. 

Our first Distance Postulate is related to the idea that any stick can 
serve as a unit of distance. We mi^it develop a formal geometry with 
one distance f miction. This would be like using inches for all distances 
whether they are thicknesses of paper or distances between stars. Our 
first postulate, the Distance Existence Postulate, reveals our preference 
far recognizing the possibility of various units of distance i n our formal 
geometry. The other Distance Postulates are motivated also by phys- 
ical experiences with distance, betweenness, and separation. 

If A, B, C are three distinct eollincai points with 8 between A and 
C, then wc want the distance between A and B plus the distance be- 
tween B and C to be equal to the distance between A and C as illus- 
trated in Figure 3-3. 



14* 



a ' T c 

Figure 3-3 



If A, B, C are three noncollinear points, then we want the distance 
from A to B (the same as the distance between A and B) to be less than 
the distance from A to C plus the distance from C to B as illustrated in 



Figure 3-4. 




Figure 3-4 



100 Distance and Coordinate Systems Chapter 3 

We are now ready to State our Basic Distance Postulates in winch 
we exercise extreme care in the choice of words so that we agree on 
exactly what they mean. As always we need to know precisely what 

we are accepting without proof in the building of our formal geometry. 
We use "unique" to mean "one and only one" or "exactly one." 



Basic Distance Postulates 

POSTULATE 15 (Distance Existence Postulate) If AB is any 
segment, there is a correspondence which matches with every segment 
€D in space a unique positive number, the number matched with AJi 
being 1. 

If C and D are distinct points, there is a unique positive number 
matched with CD according to Postulate ] 5, We may also think of this 
Dumber as matched with the set (C, D). If we associate this number 
with the segment CD, we think of it as a length. If we associate the 
number with the set {C, 1?}, we think of it as a distance. Postulate 15, 
however, says nothing about the distance from C to D if C = D; in 
other words, it says nothing about the distance between a point and 
itself. We take care of this with a definition. 



Definition 3.1 The distance between anv point and itself 
is a 



Definition 3,2 

L The correspondence that matches a unique positive num- 
ber with each pair of distinct points C and D t as in Postu- 
late 15, and the number with the points C and D if 
C = D. as in Definition 3. 1 , is called the distance function 
determined by AB or the distance function based on JT& 

2. The segment AB that determines a distance function is the 
unit segment for that distance function. 

3. The number matched with C and D, as in Postulate 15, is 
the distance from C to D or the distance between C and D 
or the length of CD. 

Notation. If P and Q are any points, not necessarily distinct, then 
the distance between P and Q in the distance function based on AB is 
denoted by FQ{in AB units) or simply by FQ if the unit is understood. 



3.2 Distance 101 



Example f (Informal) Suppose AB is a segment 1 in. long, CD is a 
segment 1 ft. long, and EFis a segment & in. long. Then 

EF (in AB units) = 6 
and 

EF (in W unite) = |. 

Although Postulates 16 and 17 could be omitted and proved as the- 
orems after the Ruler Postulate is adopted, they are included in our 
formal geometry in order to simplify the development. 

POSTULATE 16 (Distance Betweenmss Postulate) Tf A, B, C 
are collinear points such that A-B-C, Lhen for any distance function 

we have AB + BC = AC. (See Figure 3-5.) 

a B c 
4 • • m ► 



Figure 3-5 

Example 2 If A-B-C, AB - 6, BC = 8 P then AC - 14. 

POSTULATE 17 (Triangfe Inequality Postulate) If A, B, C are 
noncollinear points, then for distances in any system we have AB 4- 
BC > AC (Sec Figure 3-6.) 




Figured 

Example 3 If A, B, C are noncollinear points and if BC = 10, then 

BA + AC> 10. 

Example 4 If the lengths of the sides of a triangle are x, y t z T then 
x + y > %, y + z>x, and z+ x> y. 

Suppose that FQ is a segment 1 in. long and that 1£S is a segment 
1 ft. long. Suppose that A and B are two different points and that C and 



102 Distance and Coordinate System'. Chapter 3 

D are two different points. We use distances to compare how far A is 

from B with how far C is from D, If we say that it is 3 times as far from 

A to B as from C to D, we mean that AB - 3 CD, or that ^ = 3. 

^ CD 





(See Figure 3-7.) Our experience with physical measurements tells us 
that we should get the same comparison regardless of which distance 
function we use. Thus 

.r AB (in PQ units) n . AB (in M units) rt , 

* W (in TQ units) = 3 ' *" CD (in M units) = 3 ^ 

For example, if AB = 72 (in inches) and CD = 24 (in inches), then 
AB = 6 (in feet) and CD = 2 (in feet). Thus 

AB (in inches) _ 72 _ AB (in feet) 6 

CD (in inches) "" 24 " CD (in feet) " 2 = 

This is the basis for our next postulate. 

POSTULATE 18 (Dhtance Ratio Postulate) If PQ and RS are 
unit segments and A, B t C, D axe points such that A^B.C^A then 

AB (in 2^ units) AB (in M imits) 
GO (in PQ units) "" CD (in 15 units) 

or, equivalently, 

AB (in /p units) CD (in PQ units) 
AB (in RS units) " " CD (in R5 units) " 

In making physical measurements we recognize that there are 
many accurate foot rulers. Measurements made with an official foot 
ruler and an accurate copy of one should agree. Let us see how Postu- 
late 18 is concerned with this. 

Suppose that PQ and H5 are unit segments and RS (in FQ 
units) = 1. This means that the length of RS in PQ units is 1 or, infor- 
mally, that M is a copy of FQ. If A and B are any points, it follows from 
Postulate 18 that 

AB (in TQ waits) __ AB (in A S units) 
RS (in PQ units) " RS (in M units) " 



3.2 Distance 103 



Since we assumed that RS (in FQ units) = 1 and since RS (in Jf3 
units) = 1 by the Distance Existence Postulate, it follows that 

AB (in TQ units) AB (in RS units) 



1 1 

or AB (in TQ units) = AB (in RS units). 

This proves the following theorem. 

THEOREM 3- 1 If FQ and RE are segments such that the length 
of -RS in Pp units is 1, then for all points A and B it is true that 

AB (in M units) = AB (in PQ units). 

KXERCISJiS 3*2 

1. If A and B are points and if CD is a segment, which of the following are 

necessarily true about the number AH (in CD units)? 
(a) It is a real number. 
;b) It is a positive number. 

(c) It is a nonnegaiive number. 

(d) It is an irrational number. 

2. If you know that A and B are distinct points and that W> is a segment, 
what can you say about the number AB (in RS units)? 

3. If you know that A-B-C, what can you say about the number 
AB + EC , 

AC 

4. If you know that A, B, C are noncoHinear points, what can you say about 
the number AB + BC ? 

5. If you know that A, B, C are points, that A # G* and that FQ and H5 
are segments, what can you say about the numbers 

AB (in FQ units) , AB (in RS units) . 

AC (in FQ units) ! AC (in RS units) 

Why is it not necessary to say A ^ C in Exercise 4? 

6. If you know that A, B, C are noncoUinear points and that PQ and H5 
are segments, how does the number 

AB (in FQ units) + BC (in FQ units) 

\C (in FQ m.iiHi 

compare with the number 

AB (in H5 units) + BC (an BS units) a 
AC (in RS units) 



104 Distance and Coordinate Systemi Chapters 

7. Let A, B> C, D, E be distinct points ordered on a line I as shown in the 
figure, andjttjually spaced so that AB (in AB units) =: BC (in Tfil units) 
= CD (ia AB units)_= D£ (in ZB units). Find the following: AB (in 
BC units), £C (in BC units), CD (in BC units), and DE (in BC units). 

ABODE 
4 . . . « . +t 



8. Given the situation of Exercise 7, find the distances AB t BC, CD, DE, 

all In AC units, 

», Given the situation of Exercise 7, And the numbers AB (hi AB units), 
AB (in AC units), AB (is Bl? units), and AB (ii i i 7 ! i 1 nits). 

10. Civen the situation of Exercise 7, find AD (in A"B units), AD (in BC 
units), AD (in AC units), AD (in BE units), and AD (in AE units). 

1 1. Given the situation of Exercise 7, copy and complete the following 
proof that BD (in JU units) = 1. 

Proof: Expressing ail distances in ACT units we have the following: 

1. BD — BC + CD 1. Distance Between ness Postulate 

CD 

2. -jg = 1 2. Exercise 7; Distance Ratio Postulate 

3. CD = AM 3, Multiplication Property of Equality 

4. BD = BC + AB 4. Steps i, 3; Substitution 

5. BD = AB + BC 5. 
a AC = AB + BC 6. Q] 
7, BD = AC 7. 
& AC = 1 8. [7] 
9. BD = 1 9. 

12. In the Distance Ratio Postulate, take PQ —CD and M = JB. Using 
this special case of the postulate, prove that 

AB (in CD units) - CD (in AB units) = L 

(Does this result seem reasonable? Think of AB as a stick 2 ft. long and 
CD as a stick 3 ft. long. Then AB (in CD units) = |, CD (in AS units) = 

1 1 



Land*- 4 =10 



13. Given A, B, C, D such that AB (in C~D units) = % find CD (in AB units). 
(Does your answer seem reasonable? The given information means that 
AB is — as long as CD. Your answer means that C~D is how many times 
as long as AB?) 

14. challenge FfioBLEM. Given segments AB, CJ5» EF, prove that 

EF (in AB units) - EF (in CD units) • CD (in JF units). 

(A simple example is: The number of feet from E to F equals the num- 
ber of yards from E to F times the number of feet in a yard.) 



3.3 Une Coordinate Systems 105 

15. {An Infortnal Geometry Exercise.) fat the figure, suppose that the part 
of I between F and M is on the edge of a ruler. On this ruler is assigned 
to P and 1 to Q. What numbers should be assigned to C, D t E, and \f? 

p q CD KM 

4 — * ir « — r 9—9 — M 

16. (An Informal Geometry Exercise.) Two number lines / and m are 
placed parallel to each other as shown in the figure below. What num- 
bers should be assigned to S and T on IP What numbers should be as- 
signed to F, Q, and H on m? 

A 









s 


T 


-I 
4 




p 

—m — 


2 

Q 

•— 


V 

m 





100 
3 

a 4 1 

17. Refer to Exercise 16. If A on if is directly alxrve B on m and if 100 is as- 
signed to A, what numl>er is assigned to B? 

18. challenge problem. Hefer to Exercise 1 6. If a number x is assigned 
to a point on land if the number x' is assigned to the point directly be- 
low it on m f express x' in terms of x. 

19* In the Distance Hah'o Postulate we stated that the two equations were 
equivalent, (a) Derive the second equation from the first one using some 
of the properties of equality, (b) Derive the first equation from the sec- 
ond one using some of the properties of equality. 



3.3 LINE COORDINATE SYSTEMS 

A key feature of a number line is that different points are matched 
with different numbers. In fact, we consider al the points of the line as 
matched with all the real numbers. Such a matching is called a one-to- 
one correspondence between the set of all points on the line and the 
set of all real numbers. 

Another key feature of a number line is that the distance between 
any two points is the absolute value of the difference of the numbers 
matched with those two points. A unit segment for these distances is 
the segment whose endpoints arc matched with and 1. 

Example '" Figure 3-8, 

AB= 1-1 -(-2)| = 1. J BC=|0-(-l)| = |l| = l. 

CE = |2 - 0| = 2. AC = |0 - (-2)| = [2| = 2. 

BD = |1 - (-1)| = 2. BE = 2 - (-1)| = |3| = 3. 

AE= |2 - (-2)1, = 4. AE = |2 - <-2)| = |4| = 4, 

A B C D S 



106 Distance and Coordinate Systems 



Chapter 3 



The concept in formal geometry that corresponds to a number line 
in informal geometry is the concept of a coordinate system on a line. 
Coordinate systems are useful tools in our formal development of ge- 
ometry for planes and spaces as well as for lines. 



Definition 3,3 Let FQ be a unit segment and / a line. A 
coordinate system on I relative to PQ is a one-to-one cor- 
respondence between the set of all points of I and the set of 
all real numbers such that if points A. B, Care matched with 
the real numbers a* b t c, respectively, then 

1. B is between A and C if and only if b is between a and c and 

2, AB (in TQ units) = \a - b\. 

Definition 3,4 

2* The origin of a coordinate system on a line is the point 
matched with 0, 

2. The unit point is the point matched with 1. 

3. The number matched with a point is its coordinate. 



These definitions describe and help us to think about a coordinate 
system. But a very important issue must not be overlooked here. How 
do we know there are things such as coordinate systems in our formal 
geometry? To answer this question we return to our experiences with 
number lines and physical measurements. 

Let us suppose that TQisa. unit segment, say a segment 1 cm, long, 
and that A and B are any two distinct points on I. From our experience 
with physical rulers we know that we can lay a ruler graduated in centi- 
meters alongside I and measure distances starring from A and extend- 
ing toward B as shown in Figure 3-9. 



♦I 



Ffaove 3-D 



l' entire cere 



Of course, we can just as well start at B and measure toward A as 
shown in Figure 3-10, These ideas suggest our next postulate. 



-*/ 



Figure 3-10 



Centimeters 



3,4 Rays, Segments, and Coordinate* 107 

POSTULATE 19 (Ruler Postulate) If AB is a unit segment and 
if P and Q are distinct points on a line l t then there is a unique coordi- 
nate system on / relative to AB such that the origin is P and the coordi- 
nate q of Q is ft positive number. (See Figure 3-11.) 



I 



q(q>0} Figure 3-11 

We are now ready for a theorem. 

THEOREM 3,2 (The Origin and Unit Point Theorem) If P and 
Q are any two distinct points, then there is a unique coordinate 

system on PQ with P as origin and Q as unit point. 

Proof: 

1. There is a unique coordinate system 1. Ruler Postulate 

on PQ relative to TQ t with the coor- 
dinate of P equal to and the coor- 
dinate q of Q a positive number. 

2. PQ (in Pp units) - q - = q. 2, [T] 

3. PQ {in Pt> units) si 3. [7] 
4 q = 2 4 
5. Q is the unit point. 5, [7] 

3,4 RAYS, SEGMENTS, AND COORDINATES 

I^t A and B be two distinct points on a line I. We know from the 
Origin and Unit Point Theorem that there is a coordinate system $ 
with A as origin and R as unit point. Let X be any point of I and let x be 
its coordinate in %, Then it follows from the definition of a coordinate 
system that A is between X and B if and only if is between x and lj 
X is between A and jB if and only if x is between and I ; B is between 
A and X if and only if 1 is between and x. (See Figure 3-12,) 





X 


A 




a 








X 




A 


X 


s 






1 





A 


JC 


1 

B 


X 













I 


X 


Figure 3-12 



108 Distance and Coordinate Systems Chapter 3 

Now is between x and 1 if and only if t < 0; x is between and 1 
if and only if < % < 1; and ! is between and x if and only if .* > 1 . 
It follows from the definitions of ray, ray opposite to a rav, and segment 
that: 

ojtpAB = {X :x<0) 4 2 A i >t 



AE = {X : < x < 1} 4 ^ — G- 

A$ = {X : x > 0} 4 4 — $- 



-►/ 



1 

= {X i x < I ) < d — £ 



+f 



■w 



Similarly, if C and D are two distinct points on a line I with coordi- 
nates 2 and 5, respectively, and if X is a (variable) point on / with co- 
ordinate x f then: 

<5B = {X : 2 < x < 5} 

a {X : i > 2} 

= {X:x< 5} 

opp CD a (JE : x £4} 

opp D? = {X : i > 5} 

CD = {X : .r is real} 





.' 


£_ 




% 


2 


5 


rl 




_s 


9 




« — ' 


2 


5 


w 1 




6 


V 






2 


3 


tl 


4 — 


—A 


23 






2 


5 






C 


9 


— • , 




2 


5 


r t 




£ 


Q 










* i 



More generally, if the coordinates of two distinct points A and B 
on a line I are a and b, respectively, then it is convenient to consider 

two cases in expressing subsets of AB using set-builder symbols as 
follows. 

If <i < b If a > fr 

AB = {X;a<x<b) IB = {X : b < % < a) 



AB= {X:x> a) AB = {X : x < a] 

BX = {X : x < b) ~BA = [X:x>b] 

oppA$={X:x<a) oppA& = {X : x > a} 

oppBA = (X :x>b] oppBA={Xix<b) 



3.4 Rays, Segment*, and Coordinates 109 



EXERCISES 3.4 



In Exercises J -5, a line / with a coordinate system $ is given. The coordi- 
nates of points A,B t C arc — 3 t 0, 4, respectively. In each exercise draw the 
graph of the set and express it in terms of coordinates using a set-builder 

symbol Exercise I has been completed as i sample. 



1. BC 

Solutions BC = {X : x > 0} 



4 fc 



-3 



+ 1 



2. BA 

3. AC 

4. m 

5. £2 

In Exercises 6-10, a line / and a coordinate system § are given. In each 
exercise, given the coordinates of two points, find the coordinate of a third 
point Exercise 6 has been worked as a sample. {Note: cd A means "coordi- 
nate of A.'*) 

a cd A = 2, cd B = 5, cd X = x. If X £ ~&& and AX = 2 ■ AS, find x. 
Solution x < 2, AX = 2 - x, AB = 3. % - x = 6, x = -4 



I i ft. 



-N 



7. cd C = U cd D = 1, cd P = p. If P € c3 and ■—- = ^, find p. 

8. cdE = -5, cd F = G,cdQ = qAfQ£ oppFE and EF = FQ, findq. 

9. cd C = 29, cd H = 129, cd I = i. If GI = III, find L 

10, cd J = 15. cd K = Q> cd R = rAf JR = 2- RK T find the two possible 
values of r. 

In Exercises 11-15, given the coordinates of two points on a line, find the 
length of the segment joining these two points, 

U. cd A m 5, cd B = 173 

12. cdC= ~5,cdB= 173 

13. cdC = -& cdD = - 173 

14. cd E = 147,5, cd F = 237.6 

15. cd G = 19j, cd f* = - 17| 



1X0 Dictince and Coordinate Systems Chapter 3 

■ In Exercises 16-20, given the coordinates of two points on a line, find the 
coordinate p of a third point P satisfying the stated condition. 

16. cd A = 5, cd B = 10, P € Ail and AP = 5 

17. cdA = o,cdB = 10, P € oppBA and AP = 5 

18. «f A = 5. erf B = 10, P € BA and AP = 5 

Id. erf A = 5, erf B = 10, P € opn A^ and AP = 5 

20. cdC= -10, cdD = Q, P£ CD and PC = 17| 

■ In Exercises 21-25, given a line I and a coordinate system on it in which x 
is the coordinate of a variable point X, draw a sketch showing / and the given 
subset of/. 

21. {X : ~6<x< -2} 

22. {X :x< -6} 

23. {X : x > -2} 

24. {Xix< -6orz>; -2} 

25. {X\x< 10} 

■ In Exercises 26-35, given a tine with points and coordinates as marked in 
Figure 3-13, find the given distance. 



CQ D 


E 


F Q 


H 


P / 


JS X L MT N 


-3 t -2 


-I 


1 


2 


f 3 


4^6 G 7*8 
3 V2 6 /£ 


-V 








fl 


Figure 343 












26. CE 










31. /\) 


27. DC 










32. st 


28. HP 










33. SF 


29. IC 










34. A/g 


30, FQ 










3d. PS 



L0 



In Exercises 36-45, given a line and a coordinate system in which 
cJ A = 12, erf B = - 8 f cd C = 0, erf D = -4.2, and erf E = \M deter- 
mine if the given between ijcss relation is true or if it is false. 

36. A-B-C 41. B-E-A 

37. A-B-D 42. D-C-A 

38. D-C-E 43. D-B-A 

39. C-J5-A 44 B-E-D 

40. £-C-B 45. C-A-B 



3-5 Segment! and Ccmgrusnc* 111 

In Exercises 48-50, given that AS is a segment I ft- long and that CD is a 
segment 1 yd long, find the given distance. 

46. EF (in AS units), if EF (in CD unite) = 5 

47. CH (in W units), if GH (in AB units) = 5 

48. CD (in CD units) 

49. CD {in Al units) 

50. AB (in (715 units) 

51. Copy and complete the proof of Theorem 3.2. 

3.5 SEGMENTS AND CONGRUENCE 

In Chapter 5 we develop in considerable detail an idea called the 
congruence idea. Informally speaking, two figures arc congruent if 
they have the "same size and shape." The terms "size** and "shape" 
are not considered as a part of our formal geometry. In elementary 
geometry, we develop the concept of congruence for segments, for 
angles, and for triangles. The concept of congruence for segments is 
easy and is appropriate to include here. Intuitively, we feel that all seg- 
ments have the same shape; hence, they have the same size and shape 
if they have the same length. We make this formal in the following 
definition. 



Definition 3.5 Two segments (distinct or not) are congru- 
ent if and only if they have the same length. If two segments 
are congruent, we say that each of them is congruent to the 
other one and wc refer to them as congruent segments. 

It is convenient to have a special symbol for congruence; thus 
AB s CD means AB is congruent to (35, 

It may be helpful to compare the words "congruent and congru- 
ence" with the words "equal and equality." 

7=3 + 4 may he read as "7 is equal to 3 plus 4" 
7 — 3 -f 4 is an example of an equality. 
ABs UD may be read as "AB is congruent to (?D" 
AB as CD is an example of a congruence, 

When working with segments it is important to note carefully the 
difference between equality and congruence. The statement AB = CD 
means that A"B and UD are the same set of points, that is, that "AT?" 



112 Distance end Coordinate Systems Chapter 3 

and "CD" are different names for the same segment. This statement 
is true if and only if A = C and B = D, or A = D and B = C. (See 
Figure 3-14.) 

• • Or • • 

C D D C 

Figure 3-14 

The statement AB at UD means that AB and CD have the same 
length, that is, AB = CD, Therefore, if AB at W f then AB = CD; and 
if AB = CD, then_ AB at CD. Notethat AB = CD implies 33 « CD 
but that XB at CD does not imply AR = CD. 

We have emphasized the difference between congruence and 
equality as these concepts apply to segments. We note next some 
similarities. Recall the equivalence properties of equality, the reflexive, 
symmetric, and transitive properties. (See Section 2.8.) Congruence 
for segments has the same properties, as stated in the following 
theorem. 

THEOREM 3.3 Congruence for segments is reflexive, symmetric , 
and transitive. 

Proof; Let AB be any segment Then its length AB is a number, and 
AB = AB by the reflexive property of equality. But AB = AB implies 
that ~KB s AR Therefore congruence for segments is reflexive. (The 
remainder of this proof is assigned as an exercise.) 

Our next theorem expresses formally a simple idea concerned with 
adding or subtracting lengths. The theorem expresses this idea formally 
in terms of congruences. Theorem 3.4 is followed by Corollary 3,4.1. A 
corollary i$ a theorem associated with another theorem from which it 
follows rather easily. 

THEOREM &4 (The Length-Addition Theorem for Segments) 
If distinct points B and C are between points A and D and if 
A3 as CD, then A€ss 3D. 

Proof: There arc two possibilities as suggested in Figure 3-15. 

i— g g g a SB a 

Figure 3-15 



3.5 Segmmts and Cortgru«rtoe 113 

Case L If A-B-C, then R-C-D, and it follows from the Distance Be- 
twecnness Postulate that 

AB + BC = AC and BC + CD = BD, 

Since AS = CD (hypothesis) and since BC = BC (Why?), it follows 
from the addition property of equality that 



But 



+ BC=CD + BC. 

CD + BC = BC + CD, (Why?) 



Therefore 



AB + BC= BC+ CD (Why?) 

and AC = BD. This proves that AC ss ED, 

Case 2, If A -G-B, then C-B-D, and it follows from the Distance Be- 
tweenness Postulate that 

AC + Cfl = AB and CB + BD = CD. 

(The remainder of this proof is assigned as an exercise. Note in Case I 
that the idea, or strategy, of the proof is the addition of lengths. The 
proof of Case 2 involves the subtraction of lengths.) 

COROLLARY 3,4,1 If distinct points B and C are between 
points A and D and if AC == BD, then AB =ss £J5. 

/too/* If B and C are between A and D, then C and B are between A 
and P. The corollary follows immediately from Theorem 3.4 by inter- 
changing B and C that is, by renaming point B as point C and renam- 
ing point C as point B, 

COROLLARY 3.45 If A, B, C, D, £, £_are points such that 
A-jB-C D-E-b\ AB =s 73E, BC ssEF, then AC s 35F. 

iVoo/- Assigned as an exercise. 

COROLLARY 3A.3 If A, B s C, D, E t F are points such that 
A-B-C, Z>-£-i? AB == 0E, A^s Z5F, then BC - £F. 

Bkx>/; Assigned as an exercise, 



114 Distance a raJ Coordinate Systems Chapter 3 



We now raise a question that leads to the final theorem of ibis sec- 
tion. Suppose that a distance function determined by PQ, say, and a 

ray AB are given. Is there a point C on AB such that AC = 7? Is there 
only one such point? We know from the Ruler Postulate thai there is 

a unique coordinate system on AB relative to FQ such lhat the origin 
is A and the coordinate b of B is a positive number. Then 

a3= {X:x>0). 

(See Figure 346.) 



J L 



b * 

Figure S-IB 

This means that there is a one-to-one correspondence between the 
set of all points of AB and the set of all nonncgative numbers. There- 
fore there is exactly one point C on AB such that Lhe coordinate of C 
is Z Then 

AC = 7 - = 7. 

If D is any other point of AB, then the coordinate d of D is not 7 and 

AD = d - ^ 7, Therefore there is one and only one point C on AB 
such diat AC = 7. 

We know from our experience with rulers that we can start at any 
point on a line and 'lay off' in either direction (along either of the two 
rays on the line that have the starting point as its endpoint) a segment 
of any desired length. The example we discussed shows that in our for- 
mal geometry, given any ray, we can 'lay off," or "construct," a seg- 
ment which Is 7 units long such that the endpoint of the my is one of 
die endpoints of the segment. Of course, we could start with any posi- 
tive number other than 7 and the same reasoning would apply. We 

could start also with a given segment and "lay oF* on AB a segment 
whose length is the length of the given segment, thai is, a segment con- 
gruent to the given segment. The "lay ofF" or "construct" language is 
informal All we are really saying is that there is such a segment and 
that it is unique* These ideas lead to our next theorem. 

THEOREM 3.5 (Segment Construction Theorem) Given a seg- 
ment CD and a ray AB, there is exactly one point P on AB such lhat 

APssCD. 



3.5 Segment* and Congruence 115 

Proof: Given a ray A3, a segment CD with CD = p t and a distance 

function » there is a unique coordinate system § on AB such that A is 
the origin and such that the coordinate h of B is a positive number. (See 
Figure 3-17.) Then 

"*= iX:x>Q\, 



4 =_£ 



b P 

Figure 3-17 

Let P be the unique point of AB such that the coordinate of P is p. Then 

AP = p — = p. If Q is any point of AS other than P t then its coor- 
dinate q is different from p and 



= <j-0=q=£p. 
Therefore there is only one P on A3 such that AP = p, that is, such that 

EXERCISES 3,5 

In Exercises I~10 t a line I with a coordinate system $ is given. The coordi- 
nates of points A, B, C, D are —5, 3, 5 t 13, respectively. (Sec Figure 3-18.) 

In each exercise, determine whether the given statement is true or false. 



A 




B C 




D 


-5 




3 5 




* r 
13 

figure 3-1 


I. "KB ss CD 




6. AB=sB(7 






I.BCs^W 




7. AB= lA 






3. ATszW 




8, JB= £D 






4. AB + BC = 


AC 


9. AC nlE5 = 


riJC 




,iASUl? = 


JTC 


10. AC + HD = 


: AB + CD 


+ 2-BC 



Iii Exercises 1 1—1-5 name the property of congruence for segments which 

justifies the given statement. 

1L If AB at UD t then ITS a ATI 

12. If MssUDimdrBsiDE* thenAEsDE 

13. FQ as £P 

14. If^^^BU^CD ( andOT-EF. thenAB^EF. 

15. If XY is a segment, then XT S£ XY. 



ilfi Distance and Coordfnate System t Chapter 3 

16, Complete the proof of Theorem 3.3, 

17. Complete the proof of Theorem 3.4. 

IS. Draw an appropriate figure and prove Corollary 3,4.2. 
10. Draw an appropriate figure and prove Corollary 3.4.3. 

■ In Exercises 20-25, the coordinates of points B and C in a coordinate sys- 

— ■ 

tern on HC are given. If A-B-C and ~KE ss BC, find the coordinate of A. 

20. cdB = 0,cdC=5 23, cdB = $, cdC = $ 

21. cd B=5,cdC= 24. cd B = 159, cd C = -156 

22. cd B= ^234, cd C = - 108 25, cd B = 27, cd C = 102 



3.6 TWO COORDINATE SYSTEMS ON A LINE 

Let A and B be the origin and the unit point, respectively, of a co- 
ordinate system § on a line /, What is the coordinate of the point P if P 
is between A and B and two-thirds of the way from A to B? (See Figure 
3-19.) Obviously, adf = | since AP = § and AS = 1. 

* • ■ * M 

Figure 3-19 I 

Let C and D be points with coordinates 35.0 and 39,8, respectively, 
on a line /. What is the coordinate q of point Q if it is between C and D 
and two-thirds of the way from C to D? (See Figure 3-20.) 

G q D 
4 — » + 9—tt 

35.0 ? 39-B 

Figure 3-20 * 1 

We know that there is another coordinate system on / in which the 
coordinate of C is and the coordinate of D is 1. If Q is thought of as a 
variable point on this line with q as its coordinate in one system and % 
as its coordinate in the other system, then the particular point Q we 
want lias j as its s-coordinate. In this section, we learn how to express 
the ^-coordinates in terms of the ^-coordinates. In this example, the re- 
lation is 

q = 35.0 + 4.8*. 

Substituting § for x we get 

q = 35.0 + (4.8)(|) = 35.0 + 3.2 = 38.2, 

Therefore the point that is two-thirds of the way from the point with 
coordinate 35.0 to the point with coordinate 39.8 is the point with co- 
ordinate 38,2. 



3.6 Two Coordinate Systems on a Line 117 

The relationship between two coordinate systems on a line is useful 
in solving exercises involving points of division like the "two-thirds of 
die way from C to D" exercise as well as some exercises appearing later 
in this chapter. This relationship is also useful in gaining new algebraic 
insights. In studying equations like 



1/ — 2 -f 3x and 



Jt- 



2 1 



it is sometimes helpful to think of them geometrically in terms of co- 
ordinates on a line. 

In later chapters Wfl consider coordinate systems in a plane and in 
space. In studying a line in a plane or in space you will find it helpful 
to think of several coordinate systems associated with the line. What 
we do later is a natural extension of the groundwork we are laying in 
this chapter. 

We begin with a lemma, a "little theorem," that is useful in proving 
a "big theorem." Lemma 3.6. J plays a key role in the proof of Theorem 
3.6, which is followed by Corollary 3.6, .1. 

LEMMA 3.6.1 Let Xi and x 2 be the coordinates of distinct points 
Xi and AV respectively, on a line /. If x is the coordinate of a point 
X on /, then 

AXi X — X\ 



if X<= XiX 2 

A 2 A a % 2 — Xi 

and 

if X £ opp A'jAV 

Proof: First, suppose that X £ XiXi; then either xi < x;? or x 3 < *i- 
(See Figure 3-21.) If x t < xg, then xi < x, XX a = x - x u XgXj = 
*2 — *i, and 

AXi x — Xi 



X2X-1 


X 2 


— 


Xl 




XX, 




X 


- *1 




X2X1 




X2 


— x L 


■> 



X%Xi X2 — x% 

Xi Xz X X X 3 Xi 

• • • M 4 • • • hi 

X\ < 12 JC2 < X\ 

Figure 3-21 

If *a < xi, then x < xi, XXi = x\ - x t X 2 Xi = Xi - x 2 » and 

XX 1 X\ — X X — Xi 

X2X1 l! - X 2 X 2 - Xi 



118 Distance and Coordinate QyitNM Chapter 3 



Next suppose that X £ opp XiX 2 ; then either x x < x 2 or x 2 < x\. 
(See Figure 3-22.) If Xi < x 2j then t\ > x t XX\ = X] — x, 
X 2 Xi s %2 — Xu ^d 

\ A: 3Ci — x x — *i 

X 2 Xi *2 — *i % — a& ' 

* • • -• ► / 4 • • • ►i 

x 11 ar 2 a- 3 *! j 

x 1 < *a *a < *i 

Figure 3-22 

If xa < xi, then r > x lt XXi = x — x 1? X 2 X t = x% — x 2 , and 

XXi x — Xj x — Xi 

X 2 Xi Xi - Xa x 2 — X! ' 

Therefore, if X £ SjXj, then 

XXi x- Xi 

XgXj X2 — X] 



regardless of whether Xi < *2 or x T > x 2 . Also, if X € opp XiX 2 , then 

XX 1 x — xi 

XjjXi x 2 — Xi 

regardless of whether x-i < Xa or xi >• X2. 

THEOREM 3.6 (The Two Coordinate Systems Theorem) If Xx 
and X 2 are two distinct points of a line I, if the coordinates of Xi 
and X 2 are *i and x 2 , respectively, in a coordinate system S, andx^ 
and x 2i respectively, in a coordinate system §*, then for every point 
X on /, it is true that 

x — %\ x / — xj 



Xt — Xi x 2 — x t 

where % and x' are the coordinates of X in $ and in §', respectively. 

Proof: Suppose that we are given a line / and points X x and X 2 on I 
with coordinates Xi and x 2 in § and coordinates xi and x 2 in §' as in the 
statement of the theorem. (Sec Figure 3-23.) 

* • • 7ZI ►* 

Figure 3-23 x\ x' t {,$'} 



3.6 Two Coordinate Systems on a Line 119 



Suppose that X" £ X1X2; then it follows from Lemma 3.6.1 that 

XXj x- xi XXi xf - x^ x~ ^1 x 1 - xj 

X2Xi x 2 — xi ' X^Xi x 2 — x{ * x 2 — xi x^ — x[ 





*2 


— 






X? 


— 


*J 




*t 


— 


*f 


x 1 




«J 





Suppose that X £ opp XiX 2 ; then it follows from Lemma 3.6. 1 that 

XXj *- *i 

X 2 X t 

XXi 

x — Xi 
*2 — *1 X' 2 — X[ 

Therefore for every X on I it is true that 

x — x\ x 1 — x[ 
X'i — X\ x 2 — x\ 

and the proof is complete. 

Theorem 3.6 is closely related to the Distance Ratio Postulate. Ac- 

V X 
cording to this postulate the ratio M s independent of the distance 

XzXi 

function. Tf we use the distance ftmction of the coordinate system g in 

Theorem 3.6 y we have 

XX j |x — X\\ 



X2X1 \*2 - Xi\ 

If we use the distance function of the coordinate system §' in Theorem 

3,6, we have 

XXi = K - «;i 
x 2 x, 14-^1' 

Therefore it follows directly from the Distance Ratio Postulate that 
I x — Xi I I x? — xj I 

I Ig — *1 I I * 2 — X[ \ 

It follows from the Two Coordinate Systems Theorem that the equa- 
tion obtained from this one by omitting the absolute value symbols is 
also true. 



120 Distance and Coordinate Systems 



Chapter 3 



COROLLARY 3. 6.1 Let Xi and X 2 be the origin and unit point, 
respectively, in a coordinate system §»&. on a line L Let xi and % be 
the coordinates of X t and X 2j respectively, in a coordinate system 
§ r on /. Let k and x be the coordinates of a point X on / in the sys- 
tems %k and $ x , respectively. Then 



(i) 



and 



W 






x — X\ 

X2 — Xi 



a k t that is, x = xi + k(%2 — *i)- 



Proo/' (See Figure 3-24.) It follows from Lemma 3.6.1 that 
XX, k- 



or 



But 



XbX l 1 - 

XXi k-Q 

X2X1 = I - 



= * 



s -*. 



XX 



XjX 



L > a Why? 



Therefore 



and (!) is proved. 



K%Ki 



1 *' 




x a 




x 









1 




1 


(£*) r 


*1 




X2 




* 


CgJ 


Figure 3-24 












It follows from Theorem 3-6 that 










x - 


- xi k - 


-0 







Xi - xi 1 — 



3,6 Two Coordinate Systems on a Line 121 
Hence it follows that 

X2 — Xi 

X = *i + k(x 2 - X& 
and so (2) is proved. 

Example / (See Figure 3-25.) Think of A as X h B as X 2 . Then Xi = h 
X2 = 7, Xj = 3, *2 = 15, Then it follows from the Two Coordinate 
Systems Theorem that 

x- 1 x*- 3 



7-1 15-3 

for every point X on AB. This equation may be solved for x in terms of 
X* or for x* in terms of x. The resulting equations are 

S? ss §1? — 4 and x* = 2x + 1. 

* • • ♦ * 

S 15 ^ (S'j 

Figure 3-35 

These equations are useful in finding r' if you know x or in finding x if 
you know x\ If you are given that x' = — 6, you can use the first of 
these equations to get x = — 3j. If you are given x = 2, yon can use 
the second of these equations to getx' = 5, 

Example 2 (See Figure 3-20.) Think of P as X L , Q as X 2 . Then 
ar*-0 x-{-7) , , I. , - 

4 » ■ ■* ► 

i-7 3 (£) 

Figure 3-28 

(Or we may think of Q as Xi and P as X2. Then the theorem gives 



* -5 
0-5 
equation obtained by the first method.) 



us -£ 2. — * ^ , which simplifies to x' = \(x + 7), the same 



122 Distance and Coordinate System* Chapter 3 

Example 3 Given three coordinate systems with points and coordi- 
nates as marked in Figure 3-27, express x in terms of Jfc. Then express x' 
in terms of k and, finally, express x' in terms of *. 

ABC 

■ • • • — ► 



a s x 

o i ft ($') 



Figure 3-27 n 3 ,' (§-) 

Solution: 

(1) i-=4 * -^4 or * = ^ + 2 

w 6-2 1-0 

(2) £ - ~ n = * ~ ° or x 1 = -8k + 1 1 



6 


- 2 




1 - 


- 


x' - 


11 




Jt- 


- 


3- 


11 




1 - 


- 


x* - 


11 


_ 


x - 


- 2 



(3) sfu =tr| or ^^-zt+is 

In working an example like this one yon may need to write more 
details than we have shown. A more complete version of (3), for exam- 
ple, might he as follows: 



3- 


- 11 


6-2 




x* - 


- 11 

-8 


x-2 
4 




x' 


-11 = 


-%~ 


2) 




x* = 


_2x +4+11 




X* - 


-2ac + 


15 



Example 4 Given two coordinate systems with points and coordi- 
nates as marked in Figure 3-28, express x' in terms of x. 



4 


A B 


X 






Future 3-.2K 


3 -I 
-a -10 




IS) 
<£") 




Solutinu: 


rf - (-8) 
-10- (_8) 

3^ + 8 
-10 + 8 
x' + S 


x- 3 

-1-3 

x- 3 
-4 

x-3 







-4 



3,6 Two Coordinate Systems on a Line 123 






^_U_!9 or *=»-"> 



2 



tSxantpte 5 Given two coordinate systems with points and coordi- 
nates as marked in Figure 3-29, find x. 



B x 

-• • != ► 



10 19 (S> 

-23 -88 x (§') Figure 3-S9 



Solution: 



*-{-23) 19-0 



-38 - (-23) 10-0 

x +- 23 _ 19 

-38 + 23 B ' 10 

*+23 _ t fi 

x + 23 = -28.5 
* = -51.5 

llxamph (i A train traveled at a uniform speed on a trip from Chicago 
to New Orleans. If it was 180 miles from Chicago at 7:00 p.m. and 320 
miles from Chicago at 9:00 p.m., how far from Chicago was it at 
10:20 p.m..' 

Solution! (See Figure 3-30.) Think of hours past noon as forming one 
coordinate system and miles from Chicago another one. Suppose the 
train is x miles from Chicago at 10j hours past noon, that is, at 
10:20 PM. Then 

x- 180 10|-7 x- 180 3^ 



or 



320 - 180 9-7 140 2 



so 



x - 180 = 140(-~) = -i^, x = 180 + -292 = 413^. 
6 6 3 

1 1 g* (8) i 

180 320 x Figure 3-30 

"Hierefore the train is 413^ miles from Chicago at 10:20 p.m. 



124 Distance and Coordinate Syittm* Chapter 3 

Example 7 Given that 7. - 12, p are the respective coordinates of 
points A, B t P on a line/, that P C AB, and that AP a 3- AB, find p. 

Solution: (See Fjgu re 3-3 1 .) I n the coordinate system with A as origin 
and B as unit point the coordinate of P is 3, Then 

4 • • • hi 

p -12 7 

Figure 3-31 | 10 

Ezample 5 Given that 7, - 12, p are the respective coordinates of 
points A, B, Pon a line I, that P £ opp AB, and that AP = 3 ■ AB t find p. 

Solution: (See Figure 3-32.) In the coordinate system with A as origin 
and B as unit point the coordinate of P is —3. Then 

p - ( - 12) —3-1 p + 12 

7-(-uS = -ori- i§- = 4 - p + 12 = 7 «. p = 6 *- 
< — * — i £ M 

-12 7 p 

Figure 3-32 i o -4 

EXERCISES 3.6 

■ In Exercises 1-5, a, h, c are the respective coordinates of points A, B, C on a 
line /. State which point, A, B, or C, is between the other two, 

1. a = 0, b = 5, c = 100 

2. a = a & = 5, c = - 100 

3. a= -3,fc = -7,c = 7 

4. a = 0.500. & = 0.O50, c = 0.005 

5. a = J, h = i c = - j 

■ In Exercises 0-15. a, fc, c- are the respective coordinates of points A , B, C on 
a line /. From the given information determine whether b < c or b > & 

0, A-JB-C, o = 10, /> = 20 11. A-C-B, a = 8 t b = Q 

7. A~#-£ a = 20, b = 10 12. B-C-A, a a 8, b = 

8. B-A-C, a = 10, 6 = 20 13. lM^, a = - J, 6 = -0.66 

9. B-A-C, a = 20, h = 10 14. B-C-A, a = §> b = 0.66 
10. A-C-B, a = 0, b = 8 15, B-C-A, a m -|, h = 



3.6 Two Coordinate Systems on a Line 125 

In Exercises 16-20, a, b t c are the respective coordinates of points A, B, C 
on a line I. From the given information determine the number c. 

16. A-B-Q a = 0,fc = l,AT5s*BC 

17. A-C-B, a = 0, b = hW s*CB 
IS. B-A-C, a - 0, b = 1,AH sAC 

19. A-B-C, a = -17, b = -B,XB==B€ 
2ft A-B-C, a = -17, fc = 8, M m ~BC 

In Exercises 21-25, there is a sketch of a line with some points and coordi- 
nates marked. Find the coordinate x. 

ABC ,, R Q X 

21. i » • • ► 24. 4 • •— « ► 

I 1 m 3 13 15 (S) 

2*3 (§•) 1 x (S r ) 

X P Q A B P 

22. 4] • • • ► 25. 4 • • • ► 

x 1 (£) 16 8 30 (g) 

29 B -10 (g'J -1 8 i (g'J 

23. 4 *■ 



3 11 13 <$) 

i 1 (S) 



In Exercises 26-30, there is a sketch of a line with some points and coordi- 
nates marked. In each exercise write an equation relating x and x r and sim- 
plify it to the form x f = ax ■+■ h, (Note that a must be a number different 
from 0. For if a = 0, then x* = b and every point X would be matched with 
the same number in the system §'. Since §' is a coordinate system, we know 
tiiat different points must be matched with different numbers.) Check your 
answers by substituting the values of x at A and B in the equation to see if 
you get the corresponding values of x 1 at A and B. 



26. 4 



27. 



2S, 





A 


I 


X 




29. 




A 


X 


B 







2 


8 


X 

x' 


fin 




4 



X 

x' 



4 






i 


A 


B 




34% 




A 


B 


X 






X 

x' 


I 
& 


5 

-7 


i» ■ 




40 





400 


X 

x' 


(S) 




A 


X 


B 


















~3 
9 


X 
X* 


-10 
IB 







126 Distance and Coordinate Systems Chapter 3 

■ In Exercises 31-35, a subset S of a tine / is given in set-builder notation. 
Sketch the graph of the set and mark the ^-coordinates and ^-coordinates 
of three points of 5, Exercise 3 1 has been worked as a sample. 

3L $= {X:x=6fc + 2 l *£2} 



Sftlution 



ion: S = {x:^-2 = jt,fc<2j 



* 


£ 


1 


X 


14 


8 2 



32, S ss {X : x = 3k + 6, < k < 1} 

33, S = {X : x = -3k + B, 0<k< 1} 

34, S = {X : .*= -3* -8, * > 0} 



35. S = (X : 



3C — 3 

- 7 - : 



k-0 
I -0 



2 
14 



+ t 



, k is any real number} 



In Exercises 36-40, A and B are points on a line with coordinates 7 and 12, 
respectively. Enid the coordinate of the point F subject to the given 
condition, 

36. F C AS and AP = 3 • AB 

37. F £ AB and AP = j • A23 

38. P € opp AB and AP = 3 ■ AB 

30, P C AB and-gi- s 2 (two possibilities) 
FB 

40, F e BA and AF = ±-PB (two possibilities) 

In Exercises 41-50, A r B,P are points on a line t with coordinates 0, 1, Jt, re- 
spectively, in a system § and with coordinates —3, 7, x, respectively, in a 
system §'. Figure 3-33 is an appropriate one for Exercise 41. 



Figure 3-33 

AT 



a 



i (S) 

7 <£') 



"M 



41. II P £ AB and ^ = % find fc and *. 

AB 

42. If F £ opp AB and-— a |. find fe and k 



43. If It a -2. find 



AP 
AB 



P £ AB, or is F £ opp AB? 



3.6 Two Coordinate Systems oci a Line 127 

44, lfk = 5, find 41- 1* P € AB, or is P £ opp AG? 

AS 

45. Ifx = 17, find A- and 44- 
40. Ifx=27. find ^c and 4^- 

47. lfx = 27, find^. 

48. If* = 37, End^-. 

49. lf* = 47, find^l-. 

50. Hx = 0,find-^, 

51. Given the situation of Example 6 on page 123, express die number x 
of miles from Chicago in terms of the number t of hours past noon of 
the day the trip began. 

52. Given the situation of Example 6, find the "departure" time at Chicago. 

53. On a car trip across the country Mr. X stopped at Ridgeville and "filled 
it up"; his odometer reading was 35378. Sometime later the gasoline 
gauge read | full and the odometer read 35513. Assuming that there is 
a constant "gasoline mileage," what does the odometer read when the 
gauge reads g? 

54. Given the situation of Exercise 53, express the odometer reading m in 
terms of the amount x of gasoline in die tank, where x = 1 when the 
tank is full and x = when the tank is empty. 

55. Mr. X reeendy completed a 180-day weight reduction program. If he 
weighed 200 lb. on the tenth day and 180 lb. on the 100th day, how 
much did he weigh on the 150th day? Assume thai he loses the same 
weight each day. 

56. Given the situation of Exercise 5-5, express Mr, X's weight w [hi pounds) 
in terms of the number n of days, where n = 1 means the first day of 
the program, » = 2 means the second day of the program, and so on. 

57. Coder certain standard conditions the freezing point of water is 0° 
Centigrade and 32° Fahrenheit; the boiling point of water is 100* Cen- 
tigrade and 212" Fahrenheit. What is the temperature in degrees Cen- 
tigrade when the Fahrenheit reading is 77"? 

58. Given the situation of Exercise 57, let C denote the number of degrees 
Centigrade and F the number of degrees Fahrenheit. Obtain an equa- 
tion that relates Q and F by substituting appropriate numbers for ¥%, F*, 

P J? ("* f"* 
Cn Ca in the following equation: r ~ ' * =-?; tt • 



128 Distance and Coordinate Systems Chapter 3 

59. Starting with the equation obtained in Exercise -58, derive an equation 
in simplified form that expresses F in terms of C. 

60. Starting with the equation obtained in Exorcise 58 } derive an equation 
in simplified form that expresses C in terms of F. 

■ In Exercises 61-65, a line with points and coordinates as marked in Figure 
3-34 is given. Using the Two Coordinates Systems Theorem, x' may be ex- 
pressed in terms of x, first with A as Xi and B as X&* and then with B as Xj 

and A as Xa- The results are 

* + 1 _ * - 2 «* *> - r_5 



« i^f=T^ w 



5-2 w -1-6 2-5 

A B X 
4 f * » 1 



2 ■> x 

-1 6 x' 

Figure 3KM 

If A = A, then x = 2, r' = - I, and Equation (a) beeomes 

- y . , — - ~ n which reduces to = 0. 
o + 1 5 — 2 

61. Simplify Equation (a) if X = & 

62, Simplify Equation (b) if X' = A. 

63. Simplify Equation (b) if X = B. 

64, Tlie sum of the left members of Equations {a) and (b) h 

tf + 1 ar-^6 



6+1 -1-6 

which simplifies to 

* + i . g'-fl _ «* + i x'-e _* + i - (x- - 6) 7 . 

7 + -7 _ 7 f~" 7 = 7 = L 

Add the right members of Equations (a) and (b) and simplify. 
65. If you multiply the left member of Equation (a) by — 1 and add 1 to the 
product, the result is x + * • (-1) + L This simplifes to 

y +J . ^j , T _ -* - 1 7 
7 1 + - " 7 + 7 

_ -x + 6 x' - 6 _ x / - 6 
7 : ~7 : -1-6' 

which is the left member of Equation (b). Multiply the right member of 
Equation (a) by — 1 and add 1 to the product. Show that the result sim- 
plifies to the right member of Equation (b). 



3.7 Point* of Division 129 



3.7 POINTS OF DIVISION 

In I his section we use the Two Coordinate Svs terns Theorem b i help 
us find the coordinates of the points on a given segment which divide 
it into a given number of congruent parts or divide it in some other 
specified way. 



Definition 3.S The midpoint of a segment AB is the point 
P on AB such that 

AP = PB = \AB. 

The midpoint of a segment is said to bisect the segment or to 
divide it into two congruent parts. 

Definition 3.7 The trivection points of a segment AS are 
the two points F and Q on AB such that 



AP = PQ = QB = 

The trisection points of a segment are said to divide the seg- 
ment into three congruent parts. Similarly, points C. D, and 
E on "KB such that 

AC=CD = DE=EB = ±AB 

are said to divide AB i nto four congruent parts. This idea may 
be extended to any number of congruent parts. 



A segment AT? is a set of points. You might think of a "path" from 
A to B if you were to draw a picture of the segment. Although you may 
think of AB as a path, be careful to remember that the segment from 
A to B is the same as the segment from B to A, Indeed, AB = BA, 

Sometimes, however, we want to consider B to A as different from 
A to B. It might be helpful to think of trips and to consider the trip 
from A to B as different from the trip from B to A. The point which is 
one-third of the way from A to B is different from the point which is 
one-third of the way from B to A. This leads us to the idea of a directed 
segment. We think of a directed segment as a segment with one end- 
point designated as the starting point. A directed segment is known 
as soon as the segment is known and the starting point is known. Our 
formal definition follows. 



130 Dfst«nc« and Coordfnate Systemi 



Chapter 3 



Definition 3.8 The directed segment from A to B, denoted 
by AR is the set {Al, A}. 



It is important to note the difference between the symbol for a rav 

the symbol for a directed segment The ray symbol, as in AB a has 
a complete arrowhead, whereas the directed segment symbol, as in 

AB, has a half arrowhead. Directed segments are related to vectors, 
and half arrows are frequently used in vector notation, Vectors are verv 
useful in many branches of higher mathematics. Note that whereas 
AB = M t it is not true that AB - BA. Note that 



whereas 



= (AB,A) = {BA i A} t 



= {/U?, B} = {HA.BJ. 



Definition 3.9 Let a directed segment AB and two points 

P and Q on AB be given. If P £ AE, Q g AB y and 

AP AO 

PB ~ OB * ^ shown in ^ i § ure 3 "35, tnen ** and (J are said 

to divide AB in the same ratio, P dividing it internally and 
called an internal point of division, Q dividing it externally 

and called an external point of division. The ratio ^- is the 

PB 
ratio of division. 



A 
Figure 3-35 



f 



B 



Example 1 Given two points A and Bona line I with coordinates 4 
and 20, respectively., as indicated in Figure 3-36, find the coordinate 
of the point P on AB if P divides AB into two congruent parts. 



20 



<g> 



-► I 



Figure 3-36 



3.7 Points of Division 131 



Solution: Let x be the coordinate of the desired point P. Then 
4 < * < 20, x - 4 s 20 - x, 2x = 24, and s = 12, 

Alternate Solution: Set up two coordinate systems as indicated in 
Figure 3-37. 



P B kf 

-# •- — H 



4 * 20 (S> 



Figure $3f 



The i . 



-4 _ fe- 
-4 " 1-0 



and ar = 4+ 16A: for every P on i*. Then F divides A5 into two con- 
gruent parts if k = ^ and 

x = 4 + 16 • ^ = 4 + 8 = 12. 

Example 2 Given two points A and B o n a line I with coordinates — 3 
and 21, respectively, find the coordinates of the four points which di- 
vide it into five congruent parts. 

Solution: Let S and $' be two coordinate systems on J with coordi- 
nates of several points as marked in Figure 3-38. 



A P B 

3 ! 3! 55 — M 

* i ($'» 

Figure 3-3S 



Then 



x- (-3) _ x+ 3 __ fc-0 

21 _ (_3) :: 24 1 -0 



= Jt 



and x — 24* — 3 for every point P on /, The required points are the 
points with ^-coordinates \-> |» |, y or. in decimal form, 0.2, 0.4, 0.6 t 0.8. 
Now we compute the a:- coordinates of the division points using the 

equation x = 24k — 3. They arc 

24(0.2) - 3 = 4.8 - 3 = L8. 24(0-6) - 3 = 144 - 3 = 11.4. 

24(0.4) - 3 = 9.6 - 3 = 6.6. 24(0.8) - 3 = 19.2 - 3 = 16.2. 



132 Distant* ami Coordtnite Systems Chapter 3 

Example 3 Given two points A and B on a line I. with coordinates 3 
and -21, respectively, find the coordinates of the points P and Q on 

AB which divide AB internally and externally, respectively, in the 
ratio Z. 

Solution: Let $ and g' be two coordinate systems on I with coordi- 
nates of several points as marked in Figure 3-39. 

Q A p b 

T7. * * * ■ — H 

{&) ** 3 * -21 

Figure 349 ($') ** 0*1 

Then, as in Example 2, 

^J- = k and x s -24k + 3 
— 24 

AQ 7 
for every point F on J. Since -^ = - f A{> is less than QB and ^ is a 

point on opp AB. The coordinates x and a:' of F and Q are computed as 
follows: 



AF=k 




AQ = - ^ 




PB=1 -k 




p£ = 1 - if 




AF 7 k 




A() 7 -f 
pB 8 "' 1-tf 




FB 8 1 - k 


r- Ik s Sfc 




7 - 7Jf = -8Jf 




15* = 7 




fifs -7 




la 




**= -24(^7) + 3 = 


171 


,= _24.X + 3 = 


-8.2 






Check (using r-coordinatesK 






AF 3-(- 
PB -8.2 - 


-8.2) 
(-21) 


1L2 112 7 

" 12.8 " 1.28 " 8 




AQ 171 
Qff 171 - { 


3 
-21) 


K>S 42 7 
192 ~ 48 ~ = 8 





Example 4 Given two points A and B on a line I with coordinates 3 
and -21, respectively, find the coordinates of the points R and Ton 

AB which divide BA internally and externally, respectively, in the ratio 
-J. (Compare with Example 3.) 



3.7 Writs of Division 133 

Solution: Let $ and $' be the two coordinate systems on I with co- 
ordinates of several points as marked in Figure 3-40* Then, as in the 

preceding examples, * - +" ^ = k and x — 24fc — 21 for ever}' point 

HT 7 > 

R on /. Since ~f = ^, BT is less than 7"A and T is a point on opp BA. 
/A 8 



< i 


R 


B 


T 


4 r 

i 


X 

it 


-21 



Figure 3-40 


Then 








BR = k 






BT = - V 


RA = 1 - k 






TA = 1 - K 


BR __ 7 _ fc 






BI 7 -jtf 


flA "" 8 I - 


k 


TA" S 1 - Jf 


k-4- 

15 


(Why?) 




£ = -7 (Why?) 


x = -9.8 


(Why?) 




s? = -189 (Why?) 


Check (using 


x-coordinates). 





Bfl _ -9,8 + 21 _ 11.2 _ 112 _ 7 
RA 3 + 9.8 12.8 128 ~ 8 

BT _ -21 H- 189 ^ 168 = 1 
TA 3+189 



EXERCISES 3.7 

In Exercises 1-5, A and B are points on a line I with given coordinates a and 
&, In each exercise, find the coordinate of the midpoint of ~KB. 

L o - 5 S b = 27 4, a = S, b = -27 

2, a = -5, h = 27 $. a = 0*b= -4.8 

3. a = -5, h = -27 

In Exercises 6- JO, A and B arc points on u line / with given coordinates a 
and b. In each exercise, find the coordinates of the points which divide AB 
into the given number, n, of congruent parts. 

6. a = -3, b - - 7, n = 2 9. a = 8, b = -8, ft = 8 

7. o = 3, b = 0, « = 5 10. a = 0, b -= 10, n = 4 

8. a = -], h = 79, n = 3 



134 Distance and Coordinate Systems Chapter 3 

■ In Exercises 11-15, A and B are points with given coordinates a and b. In 

each exercise, find the coordinates of the points P and Q which divide A~B 
internally and externally in the given ratio r. 

1L a = 10, b = 20, r = I 14. a = f, b = $, r = ^ 

12. a = 20, fr = 10, r = ? 15. a = |, fc = J. r = ^ 

13. a = 26, fc = 0, f = f 

16. Prove the following theorem. 

THEOREM If tine coordinates of A and B are « and h t then the coordi- 
nate of the midpoint of AH is a + h . 

17. Prove the following theorem. 

THEOREM If the coordinates of A and B are n and h, then the coor- 
dinates of the bisection points of AB are ^ + b and - + ^ . . 

3 3 

18. A and B are points on a line / with coordinates and 1, respectively. 
Find the coordinate of the point P which divides AB externally in the 

ratio -$&, 

19. A and B are points on a line / with coordinates and 1, respectively. 

Find the coordinate of the point Q which divides AB external] v in the 
ratio -tf$L, 

20. A and B are distinct points on a line /. Is there a point P on AB*, but not 
on AB, snch that P is the same distance from A as it is from B? That is, 

is there a point P on AB which divides AB externally in the ratio y? 

■ In Exercises 2 1 -26, A and B are points on a line ( with given coordinates a 
and b, respectively, P is the point which divides AB internally in the given 
ratio r. la each exercise, select the statement from the right-hand column 
that is true. 

21. a = 0, b = 1, r = ] (A) P is between the midpoint of A~E 

and # 

22. a = 0, b = 10, r ■ { (R) AP = 1 

23. a = 10, b = 209, r = -$$ (C) P is a bisection point of AB 

24. a e 10, b = 211, r e jgj (D) BP = 1 

25. a = - 1.3, fc = 0,ra-^ (E) P is the midpoint of OT 

26. « = 0, h = 13, r = J£ (F) P is hetween the midpoint of ^ 

and A 



Review Exercises 135 

■ In Exercises 27-30, A and B are points on a line I with coordinates a and h, 
respectively. P is the point which divides A externally in the given ratio r. 
In each exercise, select the statement from the right-hand column that is 
true. 

27. a = Q.b = 1, r = ffifa (A) F-A-8 and BP = 1000 

2S. a = (Kb=z hr= jggi (B) A-R-P and HP = 0,001 

S8L Os^sUa U ^ L (C) A-B-P and BP = 1000 

30. a = 0, h = 1, r = -^ (D) P-A-tf and AP = 0.001 



CHAPTER SUMMARY 

In this chapter we have developed the concept of DISTANCE between 
two points and the concept of a COORDINATE SYSTEM on a line. We 
mtroduced five postulates: DISTANCE EXISTENCE POSTULATE, DIS- 
TANCE BETWEENNESS POSTULATE, TRIANGLE INEQUALITY 
POSTULATE, DISTANCE RATIO POSTULATE, and RULER POSTU- 
LATE, This chapter contains many definitions and theorems. A key defini- 
tion is the definition of a coordinate system. The climax of the chapter is the 
TWO COORDINATE SYSTEMS THEOREM. In Section 3.7 we applied 
the tools of this chapter to find points of division which divide a segment 
internally and externally in a given ratio. 



REVIEW EXERCISES 

In Exercises 1-5, name the property that justifies the given statement. 

13+4=4+3 

2* (AB + CD) + KF = AH + (CD + EF) 

3. If a — b> then 7? = a. 

4. (.50 + 5) • 3 = 50 ■• 3 + 5 • 3 

5. (5 + 3)-7 = (3 + 5)-7 

In Exercises 6-11, name the postulate which justifies the given statement. 

0, If P, Q, R are distinct collmear points with P-Q-R, then PQ + QR = 
PR. 

7. If X and Y are distinct points and AH is any segment, then the distance 
between X and Y in the distance function based on A2J is a positive 
number. 

8. If M, W> K arc three noncolliuear points, then RK + KW > RW. 



136 Distance and Coordinate Systems Chapter 3 

9. If A, B, C, D, E, F. G t H arc eight distinct points* then 
AB (in ~EF units) _ AB (in C?r7 nnfts)_ 
CD (in EF units) CD (in C77 units) ' 

10. If A, ii, C, D t E, F, G, Ware eight distinct points, then 

AB (in AT units) CD (in Eb units) 
AB (in C77 units) " CD (in CT units)* 

11. If A, B, C t D are four distinct points, then there is a unique coordinate 

system on A B relative to CD such that the origin is B and the unit point 
is A. 

■ In Exercises 12-20, A, B, and X are points on a line h a, h, % are their re- 
spective coordinates in one system j a\ h\ xf are their respective coordinates 
in another system. In each exercise, (a) draw and label an appropriate figure, 
(b) express x' in terms of x and simplify, and (c) express x in terms of x 1 and 
simplify. 

12. a a 0, b = 1, tf = 5, V = 8 

13. a = 0, b = 1, d = -8, V = 5 

14. c= 1, b = 0. a' = 5, tf = 8 

15. a = 17, b = 16. a' = 5, &' = 8 

16. o = 7, b = -3, ^ = 0, 6' = 1 

17. a = 0, fc = 1, a' = 0, &' = 2 

18. a = 0, b = 1, </ = a V = -1 

19. a = 0, fo = 1, a' s 1, ¥ = 

20. a = 0, /j = 100, «' = 1, fo' = 

■ In Exercises 21-30 t A, B, C f D. E, are points on a line I with coordinates 
0, 1, 2. 3, 4, respectively, in a coordinate system on I In each exercise sim- 
plify the given expression. Each expression names a number. 

21. AB (in AB units) = [T] 

22. AB (in AT unite) = \f] 

23. AC (in ATJ units) = [TJ 

24. AD (in Al units) = (TJ 

25. AD (in AC unite) s= [?] 

26. AD (in A~E unite) s [?] 

27. AE (in AB unite) = {T\ 

28. AB (in A~E unite) = 

29. Ag(inXEui U ts)_ 
AD (in AT? unite) 

30 AB (in IC unite) _ — 

" AD (in AT unite) U 



Review ExerciiB* 137 

In Exercises 31-40, A$ is a directed segment, P divides AB internally in the 
positive ratio - , Q divides A~fi externally in the ratio - , and a and h are the 

coordinates of A and B, respectively, in a coordinate system cmAB. In each 
exercise, draw un appropriate figure und find p and q, the coordinates of P 
and Q t respectively. 

31. o = 0,ft = l,f=8,ia 1 36. a = 10. fe = 5.-^ = 4 

32.a = 0=l,r:=I p *=2 37. a = -G t fc = 9, - = -i- 

i' o 

33. a = 0, h = 1, r = 2, a- = 3 38. a = -6, b = 9, - = 3 

34. a = 0. h = 1, r = 10, s = 5 39. a . = 1, b = 2, r = 5, * = 6 

35. a = 10, h = 5, J- a 4" 40. a = 1.3, & ss 7J*- a A 

s 4 * oo 

In Exercises 4 1-47, X is a point on a line / and x and k are its coordinates in 
two coordinate systems on I. The given equation tells how x and k are related 
for every X on L In each exercise, copy and complete the given statement. 

41. x = 3fe + 1 If k - 0, then x = 

42. x = 3*+i If fc= 1.thenx = |7] 

43. x = 3fc -+■ I < Jt < 1 if and only if fj] (a condition on x) 
44.x=-3Jt+l If k = -3, then a = [T| 

45. x m -3Jt +1 If Jt = -5. then x = [?] 

46. a: = — 3fc -f 1 k < — 3 if and only if JT] (a condition on x) 

47. i = — 3fc + 1 fc > — 3 if and only if \T\ (a condition on i) 




Fred Ward /Black Star 



Angles, 

Ray-Coordinates, 
and Polygons 



4.1 INTRODUCTION 

In Chapter 3 we developed definitions and postulates for the con- 
cept of distance and for coordinate system* on a line. Bctweenness for 
points is related to betweenness for red numbers through the idea of 
a coordinate system on a line. The length of a segment is related to the 
coordinates of its endpoints. 

In this chapter we develop postulates and definitions for angular 
measure or, as we usually call it, angle measure, and for ray-coordinate 
systems in a plane. Betweenness for rays is related to bctweenness of 
the coordinates matched with the rays. The measure of an angle is re- 
lated to the coordinates of the rays that form the angle. Ray-coordi- 
nates are useful in developing the properties of angles. 

Chapter 4 concludes with a discussion of polygons and dihedral 
angles. The idea of a polygon is a natural extension of the idea of a tri- 
angle, and the idea of a dihedral angle grows naturally from the idea 
of an angle. 



140 Angles, Ray-Coordinates, and Polygons 



Chapter 4 



4.2 ANGLE MEASURE AND CONGRUENCE 



sen a pie is cut into four quarters of equal size as indicated in 
Rgure 4-1, the rim is also cut into quarters. This is true regardless of 
the size of the pie. If a pie is cut in the usual manner, then with each 
piece that is less than half a pie there is an associated angle, sometimes 
called the associated central angle. This angle is the union of the two 
mys which have the point at the center of the pie as their common end- 
point and which contain the segments that are the cuts forming the 
piece of pie. For a quarter pie w r e might think of the size of this angle 




Figure 4-1 

This corresponds to thinking of a revolution as the unit of meas- 
ure. If we adopted this unit, then the measures of the angles in our ge- 
ometry would be real numbers between and j. We prefer to think of 
one revolution as equivalent to 360 degrees. Then the measures of 
angles in our formal geometry will be numbers between and 180 as 
suggested in Figure 4-2. 




Figure 4-2 

It is said that the Babylonians originated the system of measure- 
ment that is based on what we now call the degree as the unit. To them 
the stars (except the sun) appeared to be fixed on a celestial sphere 
that rotated about an axis once each day. The sun appeared to com- 



4.2 Angle Measure and Congruence 141 

plete a circular path among the stars once each year (four successive 
seasons}. They apparently knew that the length of the year was approx- 
imately 365 (lays, but, perhaps for convenience* took 360 days as their 
"calendar" year. Considering that the sun traveled over a circular path 
once each 360 days it was natural to divide that path into 360 equal 
parts and consider each part as corresponding to one day and 90 parts 
as corresponding to one season. In the early days of the Christian era, 
the Greek mathematicians of the School of Alexandria divided the 
circle into 360 equal parts and called each part a moira* This Greek 
word was translated into the Latin word de-gradus, meaning *'a grade 
or step from/' From this we get our word degree, meaning the first 
step down from a complete revolution, or -g£$ of a revolution, 

Of course* we could use other units of angular measure such as rev- 
olutions or right angles. Some units that you may not have heard of are 
mils, grads, and radians. There is no particular reason for using degrees 
for angle measures other than the fact that this is commonly done and 
lias been done for a long time. To make our development simpler, we 
base our formal geometry of angle measure on just one unit of measure, 
the degree. 

POSTULATE 20 (Angle Measure Existence Postulate) There 
exists a correspondence which associates with every angle in space a 
unique real number between and 180. 



Definition 4.1 The number which corresponds to an angle 
as in the Angle Measure Existence Postulate is called the 
measure of the angle. 



Notation. The measure of A ABC is denoted by m L ABC. 

Note that if the number of degree units in the measure of I. ABC 
is 40, then ml ABC = 40 and not mLABC = 40*. If an angle is 
marked 40° in a figure, it means that the measure of the angle is 40, 

In Chapter 3 we agreed to call two segments congruent if they have 
the same length. We make a similar agreement for angles. 



Definition 4.2 Two angles (whether distinct or not) are 
congruent angles, and each is said to be congruent to the 
other if they have the same measure. 



Notation. Z ABC s? ADEF denotes that I ABC and IDEF are 
congruent. 



142 Angles, Ray-Coordinates, and Polygons 



Chapter 4 



Although congruence and equality as applied to angles may seem 
alike, they arc in reality different ideas. It is true lhat if LA = LB, 
then LA ^ LB r but the converse is not true. Many pairs of congruent 
angles are not pairs of equal angles. Remember that an angle is a set 
of points and that two sets of points are not equal unless they consist of 
the same point*;. For the angles suggested in Figure 4-3 we have 

LABC = L CBA = LARK = L GBC = L GBK. 





Figure 4-3 

Also, LABC^ LDEF. Rut it is quite possible that mLARC = 
mLDEE If this is true, then LABC m L DEE 

The most common device for measuring, angles in informal geom- 
etry is a semicircular protractor with degree marks from to 180 evenly 
spaced on the semicircular edge. To measure an angle such as in Figure 
4-4a we can either place the protractor as indicated in (b) and read 
the measure 20 directly or vvc can place it as indicated in (c) and ob- 
tain the same measure by subtracting 65 from 85. 




Pbpus I- 4 

Another type of protractor is a circular one. (See Figure 4-5.) This 
380-degree protractor has advantages in drawing certain figures. In 
using a 360-degree protractor, as in using a semicircular one, it is pos- 
sible to obtain the measure of an angle from readings on its scale. This 
and other properties of the protractor suggest the concept of a ray- 
coordinatc system which we define later. 



4,2 Angle Measure and Congruence 143 




Figure 4-5 



EXERCISES 4.2 

1. Copy and complete the following definition of congruence of angles. 

If Z A = L B, then [fj and if »?£ A = mZB, then |TJ. 

2. (a) Use your protractor to construct three angles whose degree meas- 

ures are ©0, 90, and 135, respectively, 
(b) Why is it not possible, using our definition of angle measure, to con- 
struct an angle whose degree measure is 240? 

3. Use your protractor to find the degree measure of the angles shown. 




144 Angfas, Ray- Coordinates., and Polygon* 



Chapter 4 



4. In Exercise 3, did you find two angj.es thai arc congruent? If so, name 
them and tell why they are congruent. 

5. If Sandy Moser measures an angle and finds its. degree measure Lo be 
60 and Bob Blake measures the same angle with the same protractor 
and finds its degree measure to be 70, what can you conclude? What 
postulate are you using as a basis for your conclusion? 



Exercises 6-10 refer to the angles in Figure 4-6, 

4 



iur 



E 




U 



1KT 



rss 



R 



B 

Figure 4*0 

6. Using the notation of the figure, write two different names for equal 
angles, that is, for the same angle. 

7. Name two angles that arc congruent but not equal. 

8. Write two names of angles such that the angles named are equal (and 
therefore congruent). 

9. Name two angles that are not congruent. 

10, Can you name two angles that are equal but not congruent? 

11. Using a protractor, measure the angles of the two triangles in the fol- 
lowing figure. List those pairs of angles that appear to he congruent 




4.2 Angle Measure and Congruence 245 

12. In the figure, the readings for certain rays with endpoint V are shown 
in a circular protractor. Name four pairs of congruent angles, that is, 
name four pairs of angles such that the angles in each pair are congruent 
to each other. 




13. Without using a protractor draw six angles whose degree measures you 
would estimate to be 30, 45, 60, 90, 120* and 150, respectively. Alter 
you have drawn each angle, measure it with your protractor and see 
how good your estimate was. 

14. Draw a triangle ABC so that m Z A = 43 t mlB = 57, and m Z C = 80. 

15. Can you draw a triangle DEF so that mlD = 54, m£E = 67, and 

mZF = 70? 

16. Draw a triangle DEF so that mi. D = 54 and m/.E = 67. Use your 

protractor to find m L F, 

In Exercises 17 and 18, complete the proof of the following theorem. 

THEOREM Congruence for angles is reflexive, symmetric, and 
transitive. 

Proof; Let Z A be any angle; then in Z A is a number and mAA = m Z A 
by the reflexive property of equality. But if mLA = m/_A t then 

LA^ L A by the definition of congruence for angles. Therefore congru- 
ence for angles is reflexive. 

17. Prove that congruence for angles is symmetric 

18. Prove thai congruence for angles is transitive. 

1». (a) If LAm LB and LC^ IB, what can you conclude? 
(b) What properties justify your conclusion in (a)? 



146 Angles, Ray-Coordinates, and Polygons Chapter 4 

4.3 BETWEENNESS TOR RAYS 

Recall that if A, B, C are three distinct points on a line, then B is 
between A and C if and only if 

AB+ BC = AC 

Also, B is between A and C if and only if B is between C and A. Be- 
tweenness for points is related to betweenness for "numbers througji 
the definition of a coordinate system on a line and the Ruler Postdate. 

For three distinct coplanar rays VA t VB, VC (with a common end- 
point) we want to develop a concept of betweenness based on our in- 
tuitive notions of symmetry, our experiences with protractors, and our 
desire for additivity of angle measures in certain situations. Specifically, 

if V£J is between V'A and VC, then we want VB to be between VC and 

VA, Also, we want VB to be between VA and VC if and only if 



m 



Z AVB -r m£BVC= m I AVC 



These ideas suggest the following definition and postulate. Refer 
to Figure 4-7 us you read them. 



ix , jfr-f , J&l 



A and B are on tha / B and C are on the V A and C are on 

same aide of VC / aanw aide of VA opposite sadea of VB 

m 



V r i i» betwaan V~X and V# 



Figure 4-7 



— ♦ — ► — > — > 

Definition 4.3 If VA, VB, VC arc rays, then VB is between 

VA and VC if and only if 

1. A and B are in the same h airplane with edge VC. 

2. B and € are in the same half plane with edge VA. 

3. A and C are in opposite half planes with edge VB. 



POSTLXATE 21 (Angle Measure Addition Postulate) If VA, 

VB, VC are distinct coplanar rays, then VB is between VA and VC if 
and only if 

mLAVC = mZAVB + tnZBVC 



4.3 Betweenness for Rays 

We consider two important matters relating to Definition 4.3. 

(A) Suppose that VJ? is between VA unci VC. Is VB between VC 
and V?? 

(B) Suppose that V$ is between vX and V&, and that A', B', C 

are any points, except V,.on VA, VB, VC, respectively, as in 

Figure 4-8. Is vS' between vA' and VC?? 
In view of Definition 4,3, the questions raised in (A) and (B) amount 
to the following, expressed in terms of Figure 4-8. Suppose 
(1) A and B arc in the same 



147 




Figure 4-S 



half plane with edg<& VC, 

(2) B and C are in the same 
half plane with edge H, 

(3) A and C are in opposite 
half planes with edge VB. 

Does it follow that 

(4} C and B are in the same halfplane with edge VA, 
(5) B and A are in the same halfplane with edge VC, 
(0) C and A are in opposite halfplancs with edge VB? 

Does it follow that 

(7) A' and B' are in the same halfplane with edge VC, 

(8) & and C are in the same halfplane with edge VA\ 

(9) A' and C are in opposite halfplanes with edge VB'? 

Since (1) implies (5) and (7), (2) implies (4) and (8), and (3) implies 

(6) and (9), it follows that the answer to questions (A) and (B) is Yes. 

Thus betweenness for rays is symmetric just as betweenness for 

points is symmetric. Indeed. VB is between VA and VC if and only if 

VB is between VC and VA, and Q is between P and R if and only if Q 
is between H and P. Also, betweenness for rays depends on rays, not 
on the particular choice of points used in designating the rays. 

Postulate 21 is consistent with these properties of betweenness. 
Thus, referring to Figure 4-8. 

nUAVC = mlAVB -j- mlBVC 

mlCVA = mACVB + mlBVA 

and 

m£AVC = m£AVB + mLBVC 
mLA'VC = mlA'VB' + m£B'VC 



if and only if 

(Why?) 



if and only if 
(Why?) 



148 Angles, Ray-Coordinates, and Polygon* 



Chapter 4 



Example Consider the six coplanar and concurrent rays formed by 
the three intersecting lines in Figure 4-9. Then study the following 
two instances of betweenness in this figure and the four instances of 
"not betweenness," 




Betweenness 

1. VA is between VFand VB since (a) B and F are on opposite sides 
of VA, (b) A and F are on the same side of VB. and (c) B and A 

4 — ► 

are on the same side of VF, From Postulate 21 it follows that 
ml FVA + ml AVB = mlFVB. 

2. VA is between VB and VF. Indeed, the three statements to check 
are the same three statements that we just checked to verify 

that VA is between VFand VB, From Postulate 21 it follows that 

mlBVA + mlAVF=mlBVF, 

which should not be surprising in view of the equation 

in L FVA + m L AVB = m I FVB and the commutative prop- 
erty of addition, 

Not Betweenness 

1. VA is not between VF and VC, Why? Let us check. Are F and C 

on opposite sides of VA? Yes, Are A and F on the same side of 

VC? No. Of course, one '"So" in checking the three require- 
ments is enough to establish "not betweenness." 

An alternate method of checking betweenness in this instance in- 
volves using Postulate 21. According to this postulate, VA is between 
VF and VC if and only if 

mlFVC = mlFVA + mlAVC 



4.3 Betweenness for Ray* 149 

But this equation is a false statement. Why? It is false because the right 
side of the equation is a number, whereas the left: side is not. Indeed, 
there is no such angle as Z FVC and hence there is no number such as 
ntlFVC, (See Section 4.2.) 

2. VA is not between V? and VB. Why not? 

3. VX is not between V7?and vd, Why not? 

4. VA is not between VA and VB. Why not? 



EXERCISES 4.3 

L In the figure below at the left, explain why ST is not between BA and 
EC 





A 



2. In the figure above at the right, explain why VCis not between VA and 
~VB. 

3, In the figure for Exercise 2, is VB between VA and VC ? Is VA between 
V$and VC?? 



4. In the figure at the right, if A-B-C. 

then is BD between BA and BC? 
Why? 



D 



4 — * 



A 



G 



Exercises 5-9 refer to Figure 4-10, Assume that no two of the angles of the 

figure are congruent. In each exercise, name the missing angle. 

5. m Z AEB + ml EEC = m Z \j] 
a mlAEC + ml CRD = mZ[T| 

7. ml ABC - ml ABE = ml^} 

8. ml BED — ml\?} = mlBEC 

9. mZ[0 - mlECD = mlECB 

Flgur* 440 




ISO Angles. Ray-Coordinates, and Pofyf on* Chapter 4 

■ Exercises 10-13 refer to Figure 4- 11 . Name the missing angle or number. 



Figure 411 A 

10. mlADE + nuLEDC = mZ(7] 

11. mlDAB - nUDAC = mZ0 




12, DE + EB = JTJ 

13, AC -AE = \7] 



14 Given mlAVB = 35 
mlAVC= 115 

find mZBVCif 

(a) VB is between vA and vS, and 

(b) VA is between V7? and V?. 

15. If t in a plane, mlAVB = 70 and mlBVC = 44, find mLAVC. Is 
there just one possible value for m £ AVC? Illustrate with a figure. 

16. In the figures below. tnLABC- m/.EFG = 70 and mlDBG = 
ml HFC = 25. Prove thai LABD^ lEFll 





17, In the figures below, mlDEF s mLHKM = 120 and m/_GEF 
mlHKN = 35, Prove that IDEG s ANKM, 





18. In the figures below, mlPQS = mlDBC = m and mlSQR 
mlABD = 50. Prove that IPQR ~ ZABC. 





4.3 Betwa«nn*u for Hays 151 

Tn the figure below, mZl = mZ3 and mZ2 = mZi Prove that 
<LDEF = £DGF, 




). challenge pnoni jcm. Let VB be between VA and VC and let R<p be 

between /i? and /IS as shown in die figure below. It appears that all 
three of the following statements might be true: (a) Z CVB s Z SK^>, 
(b) Z BVA at Z ^>HP, and :c) Z C VA a / SRP. Prove that if any two 
of these three statements are true, then the third one is also true. 





Hint: There are three things to prove. 

(A) If(a)aml(b)»then(c)- 

(B) If (a) and (c), then (b), 
(Q If (b> and (c), then (a). 

Proof of {A): There is a number r such that 

m Z CVB = m / SRQ = r, Why? 

There is a number $ such that 

m / BVA = m Z £ilP = s. Why? 

Then 

miCVA = r+i= m L SRP> Why? 

Then 

ZCVAssZSRP. Why? 

Now write proofs of (B) and (C). 



152 Angles, Ray-Coordinates, and Polygon* 



Chapter 4 



4.4 RAY-COORDINATES AND THE PROTRACTOR POSTULATE 

In the same way that the Ruler Postulate provides us with a mathe- 
matical, or abstract, ruler for assigning coordinates to points, \vc want 
a mathematical protractor for assigning ray-coordinates to rays. Wc 
first define what is meant by a ray-coordinate system. Then we adopt 
the Protractor Postulate which amounts to an agreement that ray- 
coordinate systems do exist and that they are unique if we pin them 
down in certain ways. The definition is based on experiences with 
protractors. 



Definition 4A (See Figure 4-12.) Let Vbe a point in a plane 
«♦ A ray- coordinate system in a relative to V is a one-to-one 
correspondence between the set of all rays in a with endpoint 
V and the set of all real numbers x such that < x < 360 
with the following property: If numbers r and s correspond 

to rays VJR and VS in a, respectively, and if r > s, then 



mZRVS = r — * 
m£RVS = 360- (r-s) 

VR and VS arc opposite rays 



if r-s<180 
if r-s>!80 

if r - s = 180. 




Frgure4J2 



Definition 4.5 The number that corresponds to a ray in a 
given ray-coordinate system is called the ray-coordinate of 
that ray. The ray whose ray-coordinate is zero is called the 
zero- ray of that system. 



Notation. We use cd VX as an abbreviation for ray-coordinate of Va. 



4.4 Ray Coordinates 153 

Example 1 Figure 4-13 suggests a ray-coordinate system where the 
ray-coordinates of several rays are given. For this example we have: 

mAAVR = 70 - = 70 

mlAVH' = 360 - (250 - 0) = 110 

mZPVR =90-70 = 20 

mZ iW = 250 - 90 = 160 

mlFVA = 360 - (270 - 0) = 90 

VR = opp vR' since 250 - 70 s 180 

VA and VA' are opposite rays since 180 — = 180 




Figure 4-13 

POSTULATE 22 (Protractor Postulate) If a is any plane and 
VA and VR are noncolHnear rays in a, then 

(1) there is a unique ray-coordinate system § in « relative to Vin 
which cd vX = and <?<J VB = roZ A VB and 



(2) if X is any point on the B-side of VA, then cd VX (in g) = 
roZAVX 

Example 2 Figure 4-14 shows a line VA and two rays vS and V? with 
jB and X on the same side of VA.. 




V A Figure 4-14 

If ml_ AVE = 50. there is a unique ray-coordinate system in which 
cd vZ = and cd VB = 50, If mlAVX = 105 and X and B are on 
the same side of VA, then cd VX = 105. 



154 Angles, Ray-Coordinates, and Polygon* Chapter 4 

We now proceed to prove several theorems using our postulates 
aibout angles. 

THEOREM 4 J If Z A VB is any angle in a plane a and if £ is the 
ray-coordinate system in a relative to V in which cd VA = and 
cd VB = m Z -A VB, then the ray-coordinate of VX is 

(1) if VX = VZ* 

(2) 180 if f£ s= app vX. 

(3) between and 180 if X is on the B-side of VA. 

(4) between 180 and 360 if X is on the not-B-side of VA. 

Proof: Figure 4-15 consists of four parts which correspond with those 
of the theorem. 




V A 

CD 
Figure 4-15 

1. In a ray-coordinate system there is only one number matched 
with each ray. Since VX = VA and since is matched with VA. 

then is the number matched with VA. 

2. According to the definition of a ray-coordinate system two rays 
are opposite rays if and only if their ray -coordinates differ by 

180. Since cd VA = 0, since VA = opp VA, and since every 
ray-coordinate is either or a positive number less than 360, 

it follows that cd V? = 180, 

4— * > 

3. Since X is on the B-side of VA, then cd VX = mLAVX. Since 

m Z AVX is a number between and 180, it follows that cd VX 
Is a number between and 180. 

4. Let X' be a point such that VX and VX' are opposite rays. Since 
X is on the not-B-side of VA, it follows that X' is on the B-side of 
VA. Then cd VX' is a number between and 180 by part 3, and 
since cd VX is a number between and 360 which differs from 
cd VX' bv 180. it follows that cd VX is between 180 and 360. 



4.4 Ray-Coordinates 155 

The argument in part 4 expressed in symbols consists of the fol- 
lowing steps: 

0<cd VX<360 
< cd VX' < 360 
< cd V?' < 180 
cdVX- cd VX' = 180 
cd V$ = cd V?' 4- 180 
180 < cd V? < 360 

In working with a protractor we know that if we are given any an- 
gle, say Z DEF, and a ray VA on the edge of a halfpkne 3C, then we 

can draw a ray Vfi with B in 3C so that IAVB = Z DEF. Our postu- 
lates permit us to do this in. the abstract as the following theorem 
suggests. 

IHEOREM 4,2 (Angle Construction Theorem) If Z DEF h any 

angle, if VA is any ray, if 5C is any halfplane with edge VA, then 

there is one and only one halfline VR in 3C such that /.AVBst 
I DEF. 

Proof: (See Figure 4-16.) Let us suppose that I DEF, VA, and 3C are 
given as in the statement of the theorem. Let a be the plane that eon- 
tains ;1C and let F be any point in 3(1, Let § he the unique ray-coordi- 
nate system in o relative to V in which 

cd VA = and cd VP = mlAVF. 





E p Figure 4-16 

Let m Z DEF = b. Then < b < 180 and there is exactly one ray VB 
with B in 3C such that cd VB = b. Why? Then 

ml AVB = b = ml DEF 

and VB is the unique halfline in 3C such that LAVB^ Z DEF. 

The following theorem relates betweenness for rays and between- 
ness for coordinates'. 



156 



Angles, Ray-Coordinates, and Polygons Chapter 4 

THEOREM 4,3 If a ray-coordinate system in which cd VA = 0, 

cdVB = b,cdVC = c with c < 180 is given, then VB is between 

VA and VC if and only if b is between and c. 

Proof: Figure 4- 1 7 stipes ts the "given' ' or hypothesis of our theorem , 
We prove two things, 
L If b is between and c, then 

VB is between vZ and v£ 

2. If VJ?is between vA and v3, 
then h is between and c. 



Proof of 1: 



In-;, ,iv -t-17 



Statement 



L e < 180 

2. c > 

3, < b< c < 

4. mZAVB= 6-0 

5, m/_BVC=c-b 
a mZAVC = c - 

7, (b - 0) + (c - b) = c - 

8, ml AW + mZBVC = 
mZAVC 

9, VB is between VA and 

VC, 

Iteof-afSt 

Statement 

1. VB is between VA and vS. 

2. VA, VB, VC are distinct 
rays. 

3. B is on the C-sidc of VA 

4. 0, b* c are distinct num- 
bers. 

5. < c < 180 




1. Hypothesis 

2* All ray-coordinates are non- 
negative numbers, 

3. Hypothesis and steps 1, 2 

4. Definition of ray-coordinate 
system 

5. Definition of ray-coordinate 
system 

6. Definition of ray-coordinate 
system 

7. Properties of real numbers 

8. Substitution (steps 4, 5, 6, 
and 7) 

9. Angle Measure Addition 
Postulate 



L Hypothesis 

2, Definition of betweenness 
for rays 

3* Definition of betweenness 
for rays 

4. Definition of ray-coordinate 
system and step 2 

5. Hypothesis 



4.4 Ray-CoordtoitM 157 



6. < h < 180 

7. m£AVB = b-0 
mLAVC = c - 

8, mZAVB + mLBVC 
mlAVC 

9, 6 + mZBVC = c 

10, mZBVC>0 

11, fc < c 

12, 0<fc<c 



6. Steps 3 f 4, 5 and definition 
of a ray-coordinate system 

7. Definition of a ray-coordi- 
nate system 

8. Angle Measure Addition 
Postulate 

9. Substitution (steps 7 and 8) 

10. Angle Measure Existence 
Postulate 

11. Steps 9 and 10 

12. Steps 6 and 11 




THEOREM 4.4 (Angle Measure Addition Theorem) If distinct 

rays VB and VC are between rays VA and VD and if Z AVB s 
ZCVD, then LAVC sa LBVD. ' 



Proof: Figure 4- J 8 suggests two 
possibilities. We prove both cases 
at the same time. 



v a v a 

F*w*4-lS (.) (b) 

Suppose that VB and VC are distinct rays as in Figure 4-I8> From 
the Protractor Postulate and Theorem 4.3 it follows that there is a 

unique ray-coordinate system § such that cd VA = 0, < cd VD < 

180,0 < cd v3< cd VD, andO < cd VC < cd VB. Let cd v8 = b, 

cdVC = ccdvB = d. 

We must show that if A AVB m Z CVD t then Z AVC S Z BVD. 
Since 

mLAVB = h- = fc 

mlCVD= d- c 

m£AVC = c 
m£BVD = d-b 

our problem amounts to proving that if b = d — c._ then c =z. d — b. 
Suppose then that b = d — c. Using the Addition Property' of Equal- 
ity, we may add c — h to both sides of this equation. The result is 

which simplifies to c = d — h, the desired conclusion. 
Notice tho similarity of Theorem 4.4 to Theorem 3.4. 



158 Angles, Ray-Coordinates, and Polygon* Chapter 4 

COEOLLAR 1 4,4, I If distinct rays VB and VCare between rays 
VA and V/3 and if Z AVC as Z BVD, then I AVB m Z CVD, 

Proof: If VB and V(? are between vX and VD, then vS and VB are 
between VA and VD. The corollary follows immediately from The- 
orem 4.4 by interchanging VB and VC, that is, by renaming ray VB 
as ray VC and renaming ray VC as ray VB. 



COROLLARY 4,4,2 If VA, VB. VC. VD are distinct coplanar 

rays such that A-V-D, B and C are on the same side of AD, and 
Z AVB at L CVD, then Z AVC =s Z BVD. 

Proof There is a unique ray-coordinate system in which cd VA = 0, 

cd VB = b, cd W= c, cd v3 - 180, b^ c, < h < 180, and 
0<c< 180. Then 

mlAVB = b-0=b t 

ml AVC = c- = c, 

mLBVD- 180-6, 

mZCVD= 180 -c. 

Since Z AVB ss Z CVD, then b =s 180 - c. It follows that b + c = 
180, c = 180 - fe, ml AVC = mlBVD, and ZAVC^= ABVD. 

EXERCISES 4.4 

Exercises 1-10 refer to u ray-coordinate system in which the numbers as- 
signed to VA, V$. VC?, VD, V? ( and V?ure 0. 28, 47, 139, 263, and 319, 
respectively. In Exercises 2-9, compute the angle measures using the given 
ray-coordinates. 

1. Draw a figure to illustrate the given situation. 

2. mlBVC 6, mlFVE 

3. m/ BVF 7, mZAVE 

4. mlCVE 8. ml BVD 

5. mlDVE 9. mlCVF 

10. What can you say about rays VD and PF? 

11. Let three distinct concurrent rays VA t VB, VA' such that VA = opp VA' 
be given. Prove that 

ml AVB + mlHVA' = 180, 



4.4 Ray-Coordinates 159 

12. la the situation of Exercise 11, explain why it is incorrect in our formal 
geometry to say that 

mLAVB + mLBVA' = mLAVA'. 

Kxercises 13-19 refer to a ray-coordinate system in which the numbers as- 
signed to VA, VB, VC, and VD are 0, b t c, and d, respectively, where 
0<fr<e<ti< 180. In Kxercises 14-19, express the angle measures 
using the given ray-coordinates, 

13. Draw a figure to illustrate the given situation, 

14. mLAVC 17. m/CVD 

15. mLBVD 18. m/AVD 

16. mLBVC 19. mlAVB 

hi Exercises 20-20, the same situation as in Exercises 13-19 is given except 
that < b < c < 90 and 270 < d < 360. In Exercises 21-26, express the 
angle measures using the given ray-coordinates. 

20. Draw a figure to illustrate the given situation. 

21. mLAVC £4, m/CVD 

22. mLBVD 25. mLAVD 

23. mLBVC 26, mLAVB 

Exercises 27-38 refer to Figure 4-19. If the ray-coordinates of vX, vS, V6 t 

VD, VE, and VF are as .shown in the figure, name the theorem, definition, 
or postulate that justifies the statement in the exercise. 




U5G4 



27. VD is between VS and VE. 33. LBVC == Z D VE 

28. m Z CVE = ml CVD + 34. Z BVD m Z CVE 
m/DVE 

29. V? is between V?and VB. 35. mZ AVE = 140 - = 140 

30. mLBVF = ml BVC + 36. m I BVF = 180 - 40 = 140 

m Z CVF 

31. ml BVC = 70 - 40 = 30 37. LAVE BS Z BVF 

32. mLDVE = 140 - 110 = 30 38. LAVB B LEW 



160 Angles, Ray Coordfnales, and Polygons 

■ Exercises 39 and 40 refer to Figure 4-20. Use Corollary 4.4.2. 



Chapter 4 




i— * 

Figure 4-20 S 

39. If ZM\W at / HVS, name an angle congruent to AMVR. 

40. If /LSVNs* ZflVM> name an angle congruent to ZSVR. 

41. In the figure below, L DBA S L CBE, What can you conclude about 
Z 1 and Z2? What are you assuming from the figure about the points 

A t B t C t 0, El About the points A, B, C in particular? About D and E 

in relation to AC? 

B D 




iL 



B 



42. In the figure below, Z 1 — Z 2. What can you conclude about LBAE 
and Z CAD? Name the theorem you are using. What assumption are 
you making about the figure? 




4.5 SOME PROPERTIES OF ANGLES 

Two distinct intersecting lines form four angles as suggested by 
Figure 4-21. Two angles such as a and c or h and d t which appear "op- 
posite" each other in the figure, are called vertical angles. Two angles 
such as a and b or h and c are called a linear pair of angles. We state 
these ideas more precisely in the following definitions. 



Figure 4-21 




4.5 Some Properties of Angles 161 



Definition 4.8 Two angles are called vertical angles if and 
only if their sides form two pairs of opposite rays. 

Definition 4. 7 Two angles are called a linear pair of angles 
if and only if they have one side in common and the other 
sides are opposite rays. 



Note that two angles are vertical angles if their union is the union 
of two distinct intersecting lines. Note also that two angles are a linear 
pair of angles if their union is the union of a line and a ray whose end- 
point lies on that line* 

THEOREM 4. 5 Vertical angles are congruent. 

Proof: Let AB and CD intersect at V to form two vertical angles 

A AVC and Z BVD as in Figure 4-22. Let a be the plane determined 
by A, V, and C Let % he the unique ray-coordinate system in a relative 

to V in which cd VA = and cd VC = mL AVC. (Which postulate 
tells us there is one and only one ray-coordinate system with these 

properties?) For convenience, let cd VB = b, cd VC a c t cd VD = d. 




b Figure 4-SS 



The rest of the proof follows from the definition of a ray-coordinate 
system. 

b = 180 and d = c + 180 Why? 

mlSVD =d- b=(c+ 180) - 180 = c = mlAVC 
L AVC ^ L BVD 



Definition 4,8 Two angles (distinct or not) are complemen- 
tary, and each is called a complement of the other if the sum 
of their measures is 90. Two angles (distinct or not) are sup- 
plementary, and each is called a supplement of the other if 
the sum of their measures is 180. 



162 Angles, Ray-Coordinates, and Polygons Chapter 4 

THEOREM 4.6 If two angles form a linear pair of angles, then 
they are supplementary angles. 

Proof: Let a linear pair of angles, Z AVB and Z BVC, in plane a as 
indicated in Figure 4-2-1 be given. Let § be the unique ray-coordinate 

system relative to Vin a such that cd VA = and cd VB = m L AVB. 



4 • if- * ► 

Figure 4-23 A V C 

Then cd VJB < 180 and cd v2 = 180 

ml AVB + mlBVC = (cd VB - 0) + (180 - cd VB) = 180. 
Therefore Z. AVB and ZBVC arc supplementary. 

Notice that Exercise 11 of Exercises 4.4 is Theorem 4.6. 

THEOREM 4. 7 Complements of congruent angles are congruent . 

Proof: Let Z A and £ B be two congruent angles, let ZC'bea com- 
plement of Z A, and let Z D be a complement of ZB. Then 

mlA + niLC- 90 
-m/_8 + mlD = 90 

mZA := roZB 
r»'ZA -h mZC = mZB + mLD 

mlC = mlD 
IC^ ID 

THEOREM 4.8 Supplements of congruent angles are congruent. 

Proof: Assigned as an exercise, 

A special ray associated with an angle is its midray. Here is a defini- 
nition for a midray. 

Deftnitfon 4.9 A ray is a midray of an angle if it is between 
the sides of the angle and forms with them two congruent 
angles. A midray of an angle is said to bisect the angle; it is 
sometimes called the bisector of the angle or, briefly, the 
angle bisector. 



4.5 Some Properties of Angles 1 63 



Tt appcais from Figure 4-24 that an angle should have one and only 
one midray, This suggests our next theorem. 




Figure i-M 

THEOREM 4. 9 Every angl e has a unique midray. 

Pmof: Let IAVC as in Figure 4-25 be given. Let § he the unique 

ray-coordinate system in which cd VA a and cd VC = c 

= ml AVC. Then there is a unique ray V£? such that cd VB s b m ^ 
Why? 2 

ml AVE + m Z BVC = (fc - 0) 4- (<? - I) 

= c = mZAVC 
VB is between VA and VU. 
mZAVB = fc- 0= b 
mlBVC = c~ b=2b - h = h 
mlAVB = ml8VC 

Therefore VB is tetween VA and VC, and VB forms with V& and VC 
two congruent angles. Therefore VB is a midray of IAVC, 

We prove next that I AVC has only one midray. Suppose that VD 
is a midray of IAVC. Let cd VD = d, 

Then VD is between VA and V?. Why? 

< d < c by Theorem 4.3 

mlAVD= d^0 = d 

mlDVC= v - d 

c — d = d, c = 2d, d= 4 = b 




VD = VB Why? 

Therefore the midray is unique. 



164 Angles, Ray Coordinate*, and Polygons 



Chapter 4 



EXERCISES 4,3 



L In the figure several rays and their ray-coordinates are marked. Com- 
pute the measures of the following angles, 
(a) LAVB 



(b) 


LAVC 


(c) 


LAVH 


(d) 


LAVG 


(e) 


IAVF 


(f) 


LDVC 


(g) 


LDVB 


m 


LUVA 


9 


IDVH 


(1? 


LDVC 




:sn :i 



2, In a ray-coordinate system the numbers assigned to VA. VB, VC, VD 
are 0, 30, 130, 180, respectively, 

(a) Draw a figure to illustrate the given situation. 

(b) Compute several angle measures and use them to prove that VB is 
between VA and VC. 

(c) Is VB between VA and VD? Justify your answer. 

(d) IsVA between V/sand VC? J ustify your answer. 

(e) Is VC" between VBand V0P Justify your answer. 

3. In a ray-coordinate system die numbers assigned to VA and VB arc 

and 100, respectively. All of the rays and angles in this exercise are in 
the. plane determined by the points A, V, and B. 

(a) If m/.AVC = 50 and if B and C are on the same side of VA, find 
cd VC. 

(b) If m L AVC = 50 and if B and C are on opposite sides of VA, find 
cdvS. 

(c) If m Z BVD h 150 and if A and D are on the same side of VB, find 
cdvS. 

(d) If m / BVD b 150 and if A and D are on opposite sides of VB, find 
cd V5. 

(e) If cd VE = 200, us vS between VA and VE? 

(f) If cd VE = 200, is VA between V# and VE? 



4.5 Some Properties of Angles 165 

(g) If cd vS = 200, is Vj£ between V/l and VA? 

(h) If erf VF = 281, is VB between V# and V?? 

(i) If cd V? = 281, is VA* between vS and V?? 

(j) If cd VF = 281. is VF between VS and VA*? 

£k) Rod cd VK if VK b the midray of Z AVB. 

4« Given a ray-coordinate system in which the numbers assigned to VA, 

VN, VD, VY are a, b, c, d, respectively, let VA\ VN\ v5\ v9* t bo the 

rays opposite to VA, VN, VD, VY, respectively. Assume that <£, a < 
b<ic < cf <C 1 SO. Derive formulas in terms of ft, b, c. d for the follow- 
ing measures. 

(a) mXAVN (e) mlDVA 

(b) mlAVY (f) mZJDVY 

(c) mLAVy (g) mLDVN 

(d) mLAVU (h) mtDVN 

5, Referring to the figure, describe in your own words the following sets. 




(a) The union of VA and m (c) The union of Z AVB and Z BVC. 

(b) The union of VA and VB. (d) The union of Z AVB and / RVC. 

Referring to the figure, describe in your own words the following sets. 

(a) The intersection of VR and VF. \^p Xt 

(b) The intersection of ZiiVP and ZPVX 

(c) The intersection of Z H VP and I XVY. *_ 

(d) The intersection of LTVX and I XVP. 




7. Let AM be the edge of a halfplane £JC that contains points C and D. As- 
sume that A, C, D arc noncoilincar and that B, C, D are noneollinear. 
Describe the following sets. 

(a) The intersection of 3C and AB. 

(b) The intersection of Z CAP and 3C. 

(e) The intersection of Z CAD and AB. 

(d) The intersectiOD of Z CAD and Z CBD. 

(e) The union of 3C and /LB. 

(f; The union of 3C and Z CAD, 
(g) The union of 0C and ZAGB. 
(h) The intersection of JC and ZACJ3. 



166 Angles, Ray-CoordlnatM, and Potygorta Chapter 4 

8. Let four coplanar rays VA, VB, VC", VD he given such that VB is be- 
tween VA and VC, VC is between VB and VD, and VD = opp vl as 
indicated in the figure. Complete the following proof that 

mLAVB + mLBVC + mlCVD- 180. 




Why? 
Why? 



Proof: m L BVD = w Z BVC + mL CVD 

mLAVB + mL BVD = 180 

Therefore [TJ. 
9. Let four coplanar rays VA, VB, VC, VD be given such that vS is be- 
tween VA and vS, v8 = opp V&, and m£AV€ > mLAVB as indi- 
cated in the figure. Justify each step in the following proof that 

mLAVB + ml BVC + mZCVD = ISO. 

. e 

B 



4- 




B 



• M80 

v 

Proof: Let § be the unique ray-coordinate system in which cd VA = 

cdvts mLAVB = b < 180. 

*— * — > 

Then B and C are on the same side of Ay, I ,et cd VC = c. Then 

< c < 180, where a = mLA VC. Then < b < c < 180. Therefore 

mZ AVB + m I BVC + mL CVD 

= (b - 0) + (o - b) + (180 - c) = 180, 

10. In the figure AB and c3 intersect at O 
forming four angles, If tnLAOD = 133, 
find 

(a) ml CO A 

(b) mlBiX: 

(c) mZBOD 

11. If LA and Z B are supplementary angles and i is a number such that 
mZA = 3* + 6 and mZ J? = &r + 12, find the measures of Z A and 
Z B. Check your results by finding the sum of these measures. 

12. If LF and Z Q are complementary angles and y is a number such that 
mLF = y + 30 and mL{) s (/ — 30, find the measures of LF and 
Z (X Check your restilt by addition. 

13. Twice the measure of an angle is 24 more than five times the measure of 
its supplement. Find the measure of the angle and check your result. 




4.5 Some Properties of Angles 167 



14. Find the measure of an angle if its measure is twice the measure of its 

complement, 

15. Find the measure of an angle if its measure is one-half the measure of its 
complement. 

16. If / A and Z B are both congruent and supplementary, find the mea- 
sure of each. 

17. If Z A and Z B arc both congruent and complementary, find the mea- 
sure of each. 

18. The figure shows tlirec eoplanar lines and six angles marked a„ b } c t d, 
x, and y. Complete the following state- 
ments, (There are several correct re- 
sponses for some items.] 

(a) aandi? are a [3 pair of angles, 
(h) a and b arc angles. 

(c) 6 and c are \J} angles. 

(d) If a sj x, then b s* [T]- 

19. Given an angle L AVB, let 3C he the sot of all points that are on the 

same side of VB as A. that is, JC is the A-side of VB. Let X be the B-side 

of VA. Make a sketch showing JC by sliading with vertical h amines and 
X by shading with horizontal halflines. How is the interior of £AVB 
marked in your sketch? Is the interior of LAVB the intersection of 5C 
and X. or the union of 3C and X, or some other set related to 3Q and X? 

20. In the proof of Theorem 4.7, justify each of the five equations and the 
congruence. 

21. Prove Theorem 4.B. 

22. In the figure below, AC is the midray of Z BAD and VS is the midray 
of LRVT. If lBAD=z IRVT, prove that ZJ?AC= ZRV5. 






£3. Speaking informally, does the result of Exercise 22 prove that "halves" 
of congruent angles are congruent? 

24. Using Theorem 4.8 and the reflexive property of congruence for angles, 
write an alternate proof of Theorem 4.5. 

25. In the figure at the right, 

Z2=s Z3. 



Prove that 
Z1^Z3 



Z4^ Z5. 




168 Angles, Ray-Coordinates, and Polygon* 



4.6 INTERIORS OF ANGLES 

In Chapter 2 wc defined the interior of Z ABC as the intersection 

of two halfplanes, the C-side of A B and the A-side of BC. 

Another way to think of the interior of an angle is in terms of the 
rays between the sides of the angle. From our definition of between- 

ness for rays in Section 4.3, we know that if VD is a ray between the 
sides of LAVC y then every point of VD except V (that is, the hairline 
VD) lies on the A-sidc of VC and on the C-side of AV. Hence VD is con- 
tained in the interior of the angle. What we have shown for VD is true 

for every ray between VA and VC. For convenience let (R denote the 

union of the interiors of all rays between VA and VC and let $ denote 
the interior of / AVC. We have shown that (Ft lies in 3 t that is, that 

(Red. 

In Lemma 4. 1 0. 1 we shall show that if a point is in §, it is also in (ft. 
Then we shall have (R C d and £ C (R, and hence $ =s (R, which we 
state formally in Theorem 4.10. 

LEMMA 4.10.1 If a point is in the interior of an angle, then it is 
an interior point of a ray between the sides of that angle. 

Proof: Let D be a point in the interior of Z.AVC, and let a be the 

plane containing the points A, V, C, D. Then it follows from the defini- 
tion of the interior of an an^le that (1) A and D arc on the same side of 

VC and (2) D and C are on the same side of VA. We shall show diat VD 

— ? > 

is between VA and VC y and hence that D is an interior point of a ray 

between the sides of L AVC, 

Let § and §' be ray- coordinate systems in a relative to V with co- 
ordinates as indicated in the table and such that <C c <C 180. 
< a' < 180. Since < c < 180 and since C and D are on the same 

side of VA, it follows that < d < 180. Since < d < 180 and since 

A and D are on the same side of VC, it follows that < & < 180. 

S 3L 
cdVA a' 
cdVD d d' 



ud VC? 



c 



4.6 Interiors ot Angles 169 

Suppose that c = d. Then d' = (Why?), D lies on VC t and D is 
not in the interior of Z.AVC. Since this contradicts the hypothesis of 
the theorem, it follows that c^d. 

Then < d < c < 180 or < c < d < 180. Suppose that 

< c < d < 180. 

Then it follows from Theorem 4.3 that VC is between VA and VD and 

from Definition 4.3 that A and D are on opposite sides of VC. However, 

A and D are on the same side of VC, (See (1) in proof.) Since this con- 
tradicts the Plane Separation Postulate, il follows that 

< d < c < 180. 

Using Theorem 4*3 and Definition 4.3 again, we deduce that YD is 

between VA and VC and hence that D is an interior point of a ray be- 
tween the sides of £ AVC, 

We have proved the following theorem, 

THEOREM 4.10 The interior of an angle is the union of the in- 
teriors of all rays between the sides of the angle. 

Another way to consider the interior of an angle is in terms of the 
segments whose endpoints he on the sides of the angle. (Sec figure 
4-26.) 




Figure 4-26 

Given Z ABC, let D be any point on BA except B and let E be 

any point on BC except B< Let P be any interior point of the segment 
DE. In other words, P is any point between D and E. Since all of DE 

except E lies on the A-side of BC and all of DE except D lies on the 

C-side of AB t it follows that Plies in the interior of A ABC. Thus we 
have the following theorem. 

THEOREM 4.11 If AB is a segment joining an interior point of 
one side of an angle to an interior point of the other side, then the 
interior of AB is contained in the interior of the angle. 



170 Angles, Ray-Coordinates, and Polygons 



Chapter 4 



4.7 ADJACENT ANGLES AND PERPENDICULARITY 

A linear pair of angles is a special case of a pair of coplanar angles 
having one side in common and interiors which do not intersect. (See 
Figure 4-27.) It is convenient to introduce a special word for angles 
with th is property. 



4 

Figure 4-27 

Definition 4.10 Two coplanar angles are adjacent angles if 

they have one side in common and the intersection of their 
interiors is empty. 



"fl 





Figmc 4-2S 

In Figure 4-28, L AVE and Z B VCare adjacent angles and Z AVE 
and Z AVC are adjacent angles. Although Z DWE and Z EWF arc 
adjacent angles, note that Z DWE and Z DWFaxe not adjacent angles. 

All of you have a background of experience with right angles, per- 
pendicular lines, acute angles, and obtuse angles. Following are the 
formal definitions for these terms. 



Definition 4.11 An angle whose measure is 90 is a right 
angle. An angle whose measure is less than 90 is an acute 
angle. An angle whose measure is greater than 90 is an obtuse 
angle, (See Figure 4-29.) 



Sight EUjgle 
Figure 4-2M 



AouU- nriKle 



Oi'LusL' im^Uti 



4.7 Perpendicularity 171 

THEOREM 4. 12 If the two angles in a linear pair are congruent, 
they are right angles. 

Frew?/: (See Figure 4-30.) Let the measure of each angle in the linear 
pair be r. It follows from Theorem 4.6 that r + r = ISO, Therefore 
r = 90 and each of the angles is a right angle. 



Figure 4-30 

THEOREM 4.13 Any two right angles are congruent. 

Proof; Every right angle has a measure of 90. Hence all right angles 
have the same measure and hence all right angles are congruent to each 
oilier. 



Definition 4.12 If the union of two intersecting lines con- 
tains a right angle, then the lines are perriendicular. 



If Z A VB is a right angle, then AV and VB are perpendicular 

(See Figure 4-31.) 



mcs. 



• A 



~V 



A 



B' 



>A' 



Figure 4-31 Figure 432 

It b easy to show that two perpendicular lines form four right 

angles. Let A A' and BB' be perpendicular lines which intersect at V 
as indicated in Figure 4-32 and let L AVB be a right angle. Then 
LA'VW is a right angle since LA'VB' and Z AVB arc vertical angles 
and vertical angles are congruent. Also, L AVB r is a right angle since 
I AVB' and A AVB form a linear pair. Then ml AVB = 90, 
mLAVB + ml AVB' = 180, and therefore ml AVE' = 90, Why is 
Z A' VB a right angle? 



172 Angles, Ray-Coordinates, and Polygons 



Chapter 4 



When are two rays perpendicular? It seems natural enough to say 
that two rays are perpendicular if the lines which contain them are per- 
pendicular. We extend this idea to any combination of segments, rays, 
and lines in the following definition. 



Definition 4. 13 Two sets, each of which is a segment, a ray, 
or a line, and which determine two perpendicular lines are 
called perpendicular sets, and each is said to be perpendicu- 
lar to the other. 



Notation. We write © _L (B to mean that <2 and (B arc perpendicular 
sets, Wc may read <$. _L (B as "'d is perpendicular to (&' f Note that if 
a 1 (B, then OS 1 4 

If d and (B are perpendicular sets, then & is contained in some line 
I, (B is contained in some line m, and I JL m. Since perpendicular lines 
are distinct intersecting lines, it Follows that if Q, _L (B. then d and (B 
have at most one point in common. As indicated in Figure 4-33, the 
intersection of two perpendicular sets is a set consisting of at most one 
point. 




A ray porpoodieular to a 
line; intersection is empty 



ray perpendicular tn n 

ray; interaeclion \s hmi;!v 



Figure 4-33 



4.7 Perpendicul»rtty 173 

In a figure two perpendicular sets may be marked with a little 
"square comer" as illustrated in Figure 4-34. 





3. 



Figured 

A typical exercise in informal geometry involves constructing a per- 
pendicular to a line at a point on it. Our formal geometry is sufficiently 
developed now so that we can prove that such perpendiculars exist. 

THEOREM 4. 14 For each point on a fine in a plane, there is one 
and only one line which lies in the given plane, contains the given 
point, and is perpendicular to the given line. 

Proof: See Figure 4-35. Let P be a point on line I in plane a. Let 0C 
he one of the two halfplanes in a with edge I. There exists a point A 
different from P on /, Why? 



B 



X 



• R 



Figure 4-io 

Let R be any point in 3C. Let § be the unique ray-coordinate 
system in a relative to P in which cd PA = and cdPR = m/ APR. 

Let PR be the unique ray with cd T3 = 90. LAPS is a right angle 
andPB± PA. Why? 

Therefore there is at least one line in a through P perpendicular to t. 

Let m be any line in a through P and perpendicular to I Then 
m U /contains four right angles. One of these right angles is the union 

of FA and a ray, say PC, such that PC? C 3C. 

m£APC = 90 
0<cdpd< 180 
cd PC = 90 
PC = PB Why? 

m = PE Why? 

Hence there is only one line in a through P and perpendicular to L 



174 Angles, Ray-Coordinates, and Polygon* 



Chapter 4 



In connection with Theorem 4.14 there is an interesting question 
to consider. If / is a line and F is a point not on I, is there one and only 
one line through P that is perpendicular to T? (See Figure 4-36.) Maybe 
there is no line m such that m L i. Maybe there is Just one such line m. 
Maybe there are several* We shall prove later that in our formal ge- 
ometry there is one and only one line m through P and perpendicular 
to I. The point in which m intersects I is called the foot of this perpen- 
dicular from P to I. 



p 



+i 



Figure 4*36 

It is important to emphasize that all lines considered in Theorem 
4.14 lie in one plane. Tf we remove this restriction, it seems reasonable 
that there are many lines perpendicular to a given line at a given point 
on it, as suggested in Figure 4-37. This idea comes up again in a later 
chapter. 




Figure +a? 



EXERCISES 4,7 



Let a ray-coordinate system § in a plane n relative tea point Vbe such that 

the ray-coordinates of VA, VB, VCarc a, b> c, respectively- In Exercises 1-5, 
given the values of a, 6, G, determine whether or not B is in the interior of 
IAVC. 



I a = t b = 90 t c = 175 

2. a = 0, b a 90, c = 185 

3. a = 0,b = 270, c = 185 



4. a = 0, h - 270, c = 175 

5. a = 90, h = 260, c = 



4.7 Perpendicularity 175 

In Exercises 6-16, A t B, Care three noncollinear points and D is an interior 
point of BC. 

6. Is D an element of the interior of Z J9AC? 

7. Is D an element of the interior of Z BCA? 

8. Is D an element of the interior of / CAB? 

9. Is J? an element of the interior of Z CAU? 

10. Is C an element of the interior of Z DAB? 

11. Is AB a subset of I ABC? 

12. Is AB a subset of I ABC? 

1 3. Is the interior of A~fi a subset of Z ABC ? 

14. Is SUa subset of /.ABC? 

15. Is AD a subset of die union of Z.BAC and its interior? 

16. Answer each question under "reason" with a definition, a postulate, or a 
theorem in the following proof that mZ DAC + mLDAB = m£BAC. 



atcment 



(a) D is an interior point of BC. 

(b) D is in the interior of 
ABAC. 

(c) AD is between A~B and AC. 

(d) mLDAC + mLDAB = 
mLBAC 



(a) Given 

(b) Why? 
(e) Why? 

(d) Why? 



■ In Kxereises 1 7-22, state which of the following descriptions, (a) to (e), cor- 
rectly identifies the perpendicular sets in the given diagrams. 

(a) Two perpendicular rays, 

(b) Two perpendicular segments, 

(c) A line and a ray perpendicular to each other. 

(d) A ray and a segment perpendicular to each other, 

(e) A line and a segment perpendicular to each other. 



17. <- 



+ 



20. 



21. 



22. 



t 



\ 




176 Angles, Ray-Coordi nates, and Polygon* Chapter 4 

■ In Exercises 23-27, supply the missing word so that each sentence is Inie. 

23. If the angles in a linear pair of angles are congruent to each other, I hen 
each of them is a \T\ angle. 

24. If the measure of an angle is less than 90. then it is an \T\ angle, 

25. If the measure of an angle is greater than 90. then it is an [T| angle. 
20. If one angle of a linear pair of angles is acute, then the other one is [TJ. 
27- If A ABC is a right angle, then BA and S?are [?] lines. 

■ Let a set of coplanar rays be given as in Figure 4-38 such that m Z VAX = 

130, m/MVR = 65, m£SVM = 30, VT is the midray of ZflV?>, VS is 

between VM and VR t VR is between VS and VD. Use this information to 
find the measure of the given angle in Exercises 28-32. 

2& IRVS 

29. ZDVR 

30. ZDVT 

31. ZDVS 

32. ZrVTVf 

Figure 4-3S 

■ In Figure 4*39, A3, CD f EF are coplanar lines that intersect at V and 
mLAVC = 32.3 and m£AVF = 151.7. Use this information to find the 
measure of the given angle in Exercises 33-37. 

33. ZCVF 

34. LFVB 

35. LBVE 

36. AEVA 

37. AEVC 

Figure 4.30 

■ In Exercises 38-42, Z AVR in plane a and a ray-coordinate system in « such 

that cd VA = 270 arc given. In each exercise, supply the two numbers that 
are as close together as possible and that make the resulting statement true, 

38. If A AVB is obtuse and cd vS < 90, then cd VB is a number between 
(T|aiid{¥J. 

39. If Z AVB is acute and cd VS < 270, then cd VB is a number between 

{?] and [T]. 

40. If Z AV.fi is obtuse and cd vB > 90, then cd VB is a number between 

\T\ and [7|. 

41. If I AVB is acute and cd v3 > 270 } then cd V$ is a number betwoen 
[T]and[7]. 

42. If/ AVB is a right angle, then cd V$ is [7] or [|| 

i 




4.8 Polygons 177 



43. Copy and complete the proof of the following theorem. Draw a figure 
to help you understand the result. 

THEOREM U I A VB, Z BVC, and L CVA are three coplanar angles 
such that the interiors of no two of them intersect, then the sum of the 
measures of these three angles is 360. 

Proof: In plane A VB there is a unique ray-coordinate system § relative 

to Vin which cdvX = and cd VB = ml AVB. From the [T] Postulate 

it follows that m Z AVB is less than \J] t hence that cd VB is less than [T|. 
By JTj the interiors of A. AVB and /.AVC do not intersect. Therefore 

C and B lie on [T] (the same side, opposite sides) of AV. Therefore the 

ray-coordinate of VC, call it c> is [T] than 180. Since the interiors of 
Z BVC and Z AVC do not intersect, it follows that c is less than h + [r\. 
Then 

mLAVB= h -0 = b 

mlBVC = \2\ 

mLCVA = fT] 

mZAVB + mlBVC + mLCVA = b + (c - b) + = 30Q 

44. In the proof of Theorem 4J4 there is a statement that m J ! is the 
union of four rigjit angles. One of these angles is described in terms of 

PA arid a ray PC such that PC C 3C. Describe the other three angles 
in a similar way, 

4.8 POLYGONS 

As stated before, triangles are three- sided polygons and quadri- 
laterals are four-sided polygons. A polygon is a plane figure with n ver- 
tices and n sides, where n is an integer greater than or equal to 3, The 
figures shown in Figure 4-40 are polygons, but the figures shown in 
Figure 4-41 are not. 






Ffpir»4-4J 



178 Angles, Ray-Coordinates, and Pelyfont 

Our formal definition is as follows, 



Chapter 4 



Definition 4.14 Let n be any integer greater than or equal 
to 3, Let Pi, P 2 . . . . , P n u P* he n distinct coplanar points 

such that the n segments P,P 2l P^, P^-iA, Jyi have 

the following properties: 

1. No two of these segments intersect except at their 
endpoints. 

2. Xo two of these segments with a common endpoint arc 
collinear. 

Then the union of these n segments is a polygon. Each of the 
n given points is a vertex of the polygon. Each of the n seg- 
ments is a side of the polygon. 



Tf n — 3, the definition of a polygon yields a triangle; if n — 4, it 
yields a quadrilateral. Sometimes a polygon with n sides is called an 
n-gon. For example, a polygon with 13 sides is a 13-gon. The following 
list gives the names commonly used for the polygons having the num- 
ber of sides indicated. You should learn these names. 



• umber of Sides 



N 



ante 



10 

12 



'triangle 

Quadrilateral 

Pentagon 

Hexagon 

Heptagon 

Octagon 

Decagon 

Dodecagon 



Notation. The polygon whose vertices are Fi, F& . . . a P„ and whose 
sides are TJ^ t Pp^, .... P« ^.P^is called the polygon PjP 2 . . . P„. 



Definition 4. 15 Two vertices of a polygon that are end- 
points of the same side are called consecutive vertices. Two 
sides of a polygon that have a common endpoint arc called 
consecutive sides. A diagonal of a polygon is a segment 
whose endpoint s are vertices, but not consecutive vertices, 
of the polygon. 



4.8 Polygons 179 



Most of our work with polygons is restricted to polygons like those 
ill Figure 4-42, 

Figure 4-4£ 

We usually exclude polygons like those in Figure 4-43 because they 
are not convex polygons. 









Figure 4-43 

The polygons that interest us are the convex polygons, The follow- 
ing is our formal definition. 



Definition 4. 16 A polygon is a convex polygon if and only 
if each of its sides lies on the edge of a half plane which con- 
tains all of the polygon except that one side. 



You in ust be careful not to confuse the idea of a convex polygon 
with that of a convex set as defined in Section 2.5. A convex polygon 
is not a convex set of points, although the union of a convex polygon 
and its interior is (sec Definition 4,17). 

Figure 4-44 illustrates Definition 4-16, It shows the halfpkncs as- 
sociated with the sides as required in the definition. 




180 Angles, Ray-Coordinates, and Polygon* Chapter 4 

Figure 4-45 shows a quadrilateral ABCD that is not convex. Neither 

the D-side of BC nor the A-side of BC contains all points of the quadri- 
lateral not on BC. Draw a figure to convince yourself thai every triangle 
is a convex polygon. 




Ffgurc+-*5 



Definition 4.17 The interior of a convex polygon is the in- 
tersection of all of the Imlfplanes, each of which has a side of 
the polygon on its edge and each of which contains all of the 
polygon except that side. 



Figure 4-46 illustrates this definition for a convex quadrilateral 
ABCD. 

3C = C-side of A~§ 
6 = D-side of BC? 
g = A-side of cB 
DC = B-side of Ha 

Interior of ABCD = 3C (1 £ O $ PI 3C 



Definition 4, 18 An angle determined by two consecutive 
sides of a convex polygon is called an angle of the polygon. 
Two angles of a polygon are called consecutive angles of the 
polygon if their vertices are consecutive vertices of the 
polygon. 



For a triangle, any two of its vertices are consecutive vertices, any 
two of its sides are consecutive sides, and any two of its angles are con- 
secutive angles. For a quadrilateral we use the word opposite when 
consecutive is not applicable as in the following definition. 




Definition 4.19 If two sides (or vertices, or angles) of a 
quadrilateral arc not consecutive sides (or vertices, or angles), 
then they are opposite sides {or vertices, or angles) and each 
is said to be opposite the other. 




4.9 Dihedral Angles 181 
Figure 4-47 illustrates this definition. For this example we have 

A and C are opposite vertices. 
B and D arc opposite vertices, 
AB and CO are opposite sides, 
BC and 73^ are opposite sides, 
LA and / C are opposite angles, 
L B and L D are opposite angles. 

Figur* 4-47 

4.9 DIHEDRAL ANGLES 

Angles are plane figures. Every angle is a subset of a plane. Closely 
related to the idea of an angle is the idea of a dihedral angle. Sometimes 
we say plane angle when we want to emphasize that an angle is not a 
dihedral angle. A plane angle is the union of two noneollinear rays hav- 
ing the same endpoint. A dihedral angle is formed by two halfplanes 
and a line. Here is the formal definition. 



Definition 4*20 If two noncoplanar halfplanes have the 
same edge, then the union of these halfplanes and the line 
which is their common edge is a dihedral angle, The union of 
this common edge and either one of these two halfplanes is a 
Face of the dihedral angle. The common edge is the edge of 
the dihedral angle. 



A dihedral angle is suggested by Figure 4-48. In this diagram, 
B and C are poinfe on the edge, A is a point in one face but not on the 
edge, D is a point in the other face but not on the edge. A suitable 
name For this dihedral angle is A-BC-D or LA-BC~T>. The letters 
at both ends of this symbol are names of two points, one in each face 
but not on the edge; the two letters in the middle are names of distinct 
points on the edge. 



Am 




Figure 4-48 



182 Angles, Ray-Coordinates, and Polygon* 



Chaptor 4 



The same way two intersecting lines form four angles, two inter- 
secting planes form four dihedral angles as indicated in Figure 4-49. 




LAVE 
LDVC 
LCVD 
Z.DVA 




Figure 4-49 



Definition 4,21 Two dihedral angles, such as A-UV-B and 
D-IJV-C in Figure 4-49, which have a common edge ami 
whose union is the union of the two intersecting planes are 
vertical dihedral angles. 



EXERCISES 4.9 



I, Copy and complete the following definition of a pentagon: 

Let F, Q, R t S, T be five df&ttoct eopbmar points such that the five seg- 
ments [T|, EH, EH* S [D have the following properties. 

(a)[lJ 

mm 

Then the [T] of these five segments is a pentagon. 

In Exercises 2-7, one of the sides of hexagon ABCDEF is given. In each 
exercise, name two sides which arc consecutive with the given side. 

5. DE 



2. AB 

3. BC 

4. CD 



6. IF 

7. FA 



4.9 Dihedral Angles 183 




8. Copy and complete the following proof that if ABCD is a convex quad- 
rilateral, then the interior of AC is contained in the interior of the 
quadrilateral. 

Proof; Let if denote t he interior of AU a nd let $ denote the i nterior of 
the quadrilateral. 

U 



C C-side of AB 
C C-sidc of AD 

C A-side of BC 
C A-side of CD* 



But g is the intersection of [Tj. Therefore $ gj. 
9. (a) Draw a convex pentagon and all of its diagonals, 

(b) How many diagonals does a convex pentagon haver 

(c) Does the interior of each diagonal of a convex pentagon lie in the 
interior of the pentagon? 

10. Draw a pentagon which is not convex. Label it and explain why it is 
not convex. 

11. Draw all of the diagonals of the pentagon you drew in Exercise 10- 
I low many diagonals are there? 

Figure 4-50 shows a labeled cube. In Exercises 12-16, a quadrilateral whose 
vertices arc vertices of this cube is given. la each exercise, write a suitable 
name, using four of the letters from the figure, for a dihedral angle contain- 
ing the given quadrilateral. 



12. ABCD 

13. ABFE 

14. BCGF 



15, CGHD 

16. I'Mlll 




B Plgvn 450 

Exercises 17-21 also refer to Figure 4-50. Base your answers on examining 
the figure. In some cases the formal geometry that we have developed up 

to this point is inadequate to prove the correctness of these answers. 



17. Is A-BF-H a dihedral angle? 

18. Is A-BF-C a dihedral angle? 

19. Is A-BF-E a dihedral angle? 



20. Is A-BF-D a dihedral angle? 
2L Is G-HE-C a dihedral angle? 



184 Angles, Ray-Coordinales, and Polygon* 



Chapter 4 



In Exercises 22 and 23, a dihedral angle A-BC-D* a line I in plane ABC 
through A and not intersecting BC, and a line m in plane DBC through D 

and not intersecting BC arc given. 

22, Draw a figure to illustrate this situation. 

23. Explain why I and m do not intersect. 

24* Figure 4-51 shows two planes, a and & intersecting in line AB. How 
many dihedral an^es arc formed by these two intersecting planes? 
Name them. 

25. I>efine the interior and exterior 

of a dihedral angle. 

26. Define adjacent dihedral angles. 

27. Name a pair of adjacent dihe- 
dral angles in Figure 4-51. 

28. Name a pair of vertical dihedral 
angles in Figure 4-51 . 



Figure 4-51 




CHAPTER SUMMARY 

The central theme of this chapter is properties of angles. We intro- 
duced three new postulates; The ANGLE MEASURE EXISTENCE POS- 
TULATE, the ANGLE MEASURE ADDITION POSTULATE, and the 
PROTRACTOR POSTULATE. The definitions include the following. 



CONGRUENT ANGLES 
MEASURE OF AN ANGLE 
BETWEENNESS FOR RAYS 
RAY-COORDINATE SYSTEM 
VERTICAL ANGLES 
LINEAR PAIR OF ANGLES 
RIGHT ANGLE 
OBTUSE ANGLE 
ACUTE ANGUS 
COMPLEMENTARY ANCLES 



SUPPLEMENTARY ANGLES 
MIDRAY OF AN ANGLE 
ANGLE BISECTOR 
ADJACENT ANGLES 
PERPENDICULAR SEGMENTS, 

RAYS, AND LINES 
POLYGON 
CONVEX POLYGON 
DIHEDRAL ANGLE 



Review Exarclses 185 

This list of terms should remind you of many of the ideas and theorems 
of tliis chapter. Ray-coordinates have been used extensively in developing 
the ideas and in proving the theorems that are included Theorems 42 
(ANCLE CONSTRUCTION THEOREM), 4,4 (ANGLE MEASURE AD- 
DITION THEOREM). 4.5, 4.6, 4.7, and 4.8 are important for the work that 
you will do in Chapter 5. Be sure that you know die statements of these 
theorems. 



REVIEW EXERCISES 

In Exercises 1-10, write definitions for the following terms. 

1. Congruent angles 

2. Vertical angles 

3. Adjacent angles 

4. linear pair of angles 

5. Complementary angles 

6. Supplementary angles 

7. Right angle 

8. Acute angle 

9. Obtuse angle 

10. Dihedral angle 

In Exercises 1 1 -20. a, b, c are the ray-coordinates of VA, VB, V T C, respec- 
tively. In each exercise, determine if one of the three rays is between the 
other two. If one is, name it. If none is, write "none,* 1 

11. a = 0, b = 10, c = 170 

12. a = 0, b = 10, c = 180 

13. a = t b= 10, c = 190 

14. a = 0, b = 10, c - 200 

15. o = 350, h= 50, c= 110 

16. a = 270. b = 180, c = 135 

17. a = 270. b m a c = 135 
IS. a = 180, h = 90, a = 20 

19. a = 135, b = 315, c = 

20. a. = % b m 300, c = 100 



186 Angles, Ray-Coordinates, and Polygon! Chapter 4 

■ In Exercises 21-30, a and b sire the ray-coordinates of VA and VB, respec- 
tively. In each exercise, compute the measure of / AVB. 

2L a = 38, b = 106 

22. a = 300. fo = 150 

23. a = 300, b = 100 

24. a = 359, h = 1 
23. o ^ 270, h = 100 

26. a = 38, & = 50 

27. a = 198 s b = 

28. a = 15, fc = 300 

29. a = 6, b = 40 

30. a = 315, b = 345 

■ In Exercises 31-40. 

cdVA = cdvf= 150 

tfd W? a 10 cd VK - 200 

afVC?=20 cdv£ = 2S0 

ar\S = 30 crfVM=300 

crfv£=40 «/V>f=305 

cd V?=50 cdVP = 310 

erf v2 = 75 erf VQ = 315 

cd VH = 90 erf V7* = 325 

erf V?= 100 crfVS = 330 

In each exercise, determine whether or not the given statement is true. 

31. LBVffm IJVL 

32. Z7VH = LDVE 

33. Z SVB = Z UVF 

:I4, Z B VE and Z DV7J are supplementary angles. 

35. Z CVE and Z SV7 are supplementary angles. 

36. Z.KVL and LAVE are complementary angles. 

37. W? is the midray of L DVJ. 

38. V0 is the midray of L MVP 

39. V^ k the midray of Z KVE, 

40. Z SVil is an acute angle, 

41. Explain how betweenness for rays is related to betvveenness for ray- 
coordinates. 



Review Extremes 187 

42. Explain bow betweenncss for rays is related to the addition of angle 
measures. 

43. Ii LA and Z B are complementary angles and if mZA = X and 
mlB= 3* + 10, findr. 

44. If Z T and Z W are supplementary angles and if mLT = X and 
mZW=3*+ 10, find x. 

45. Use ray-coordinates to prove that vertical angles are congruent. 

46. Use ray-coordinates to prove that the angles of a linear ]>air are 
supplementary, 

47. If A-V-A' r B-V-ff, cdVA = 143, cd vS<cdVA, and tX' I «"#\ 
find ed VB, cd W, cd VB'. 

4 S. 1 f A BCD is a convex quadrilateral and £ is a point not in the plane of tiie 
quadrilateral, which of the fallowing are names of dihedral angles: 
A-BC-B, A-BC-E, D-BC-E, E-AD-C? 

49. Explain why a convex polygon is not a convex set. 

50. Explain why a dihedral angle is not a convex set. 





Chapter 




CtHirtvsy of Quebec Government TouriH Office 



Congruence 
of Triangles 



5.1 INTRODUCTION 

Suppose you placed a sheet of carbon paper between two sheets of 
paper and drew a picture of a triangle on the top sheet Will the carbon 
copy of the triangle that appears on the second sheet be the same size 
and shape as the one you have drawn on the top sheet? The idea of two 
physical objects being carbon copies of one another is what we have 
in mind when we say that the two objects have the same size and shape. 
Obviously, we cannot take a physical object such as a house and, using 
carbon paper, make a carlxm copy of it We can, however, take the 
same set of blueprints that were used in constructing one house and 
construct another house that is exactly like it, diat is, having the same 
size and shape as the first one. 

In this chapter we are concerned witii the "size and shape" of ge- 
ometrical objects such as segments, angles, and triangles. These geo- 
metrical objects are not the physical objects that we draw on our paper. 
They are the mathematical objects which exist in our minds and whose 
properties have been described in our postulates, definitions, and the- 
orems. The mathematical concept corresponding to "same size and 
shape" is congruence. We, have already defined congruence of seg- 
ments in Chapter 3 and congruence of angjes in Chapter 4. Since all 
segments have the same shape, we say that two segments are congruent 



190 Congruence of Triangle* 



Chapter 5 



if they have the same size, that is, if the measures of their lengths are 
equal. Similarly, two angles have the same size and shape or are con- 
gruent if their measures are equal 



5.2 CONGRUENCE OF TRIANGLES 

In order to arrive at a definition of congruence for geometrical tri- 
angles, let us consider again two physical triangles, one of them a car- 
bon copy of the other. Consider first a triangle that has sides of three 
different lengths, a scalene triangle. The definition we will obtain ap- 
plies, however, to all triangles regardless of their shapes. Figure 5-1 
shows two scalene triangles, A ABC and its carbon copy. AA'B'C 




If you arc to make A ABC "fit" on AA'B'C, you must match up the 
vertices of the two triangles according to the scheme in Figure 5-1: 

A <— * A', 

B «— ► B', 

C <— + C. 
This matching is called a one-to-one correspondence between the two 
sets of vertices. The correspondence between the vertices of the two 
triangles also gives a correspondence between the sides of the triangle. 



A75 
ISC 



A'W, 



If a correspondence between the vertices of two triangles is such 
that the corresponding angles and corresponding sides of the two tri- 
angles are congruent then the correspondence is called a congruence 
between the two triangles. The correspondence we have been discuss- 
ing is a congruence between A ABC and AA'&C. On the other hand, 



5.2 Congruence of Triangles 191 

it is possible to write a correspondence between the vertices of the two 
triangles that is not a congruence. For example, the correspondence, 

A « — * B\ 

B +-* C t 

C ^^ A\ 

is not a congruence between the two triangles shown in Figure 5-1 be- 
cause, by this matching of vertices, it is not possible to make A ABC 
coincide with &A'WC. Write four more correspondences between 
the vertices of A ABC and AA'FC such that no two are the same and 
such that none is a congruence between the two triangles. 
It is convenient to write a correspondence, such as 

A < — ► A\ 

B <— » W, 

C<— » C, 

on one Une as ABC ± — * A'B'C When this notation is used, it is un- 
derstood that the first letter on the left corresponds to the first letter 
on the right of the double arrow, the second corresponds to the second, 
and the third corresponds to the third as shown below. 



tf 



BC <— ► A'B'C 

V .. |» 

1 



Note that there arc several ways of writing this same correspond- 
ence, For example, both BAC 4 — * B'A'C and CBA « — » CB'A' 
name the same correspondence as ABC * — * A' WO. 

The foregoing discussion about corresponding vertices of physical 
triangles can be made to apply equally as well to abstract geometrical 
triangles. Thus, if ABC i — > DEF is a correspondence between the 
vertices of any two geometric triangles, this correspondence provides 
us with three pairs of points. We are interested primarily in these points 
as vertices of the angles of the triangles and as endpoints of the sides of 
the triangles. In connection with this correspondence we speak of six 
pairs of corresponding parts. Three of these six pairs are pairs of angles: 

LA *-^ ID, 

L B < — > L E, 

IC ^— £R 

The other three pairs are pairs of sides: 

AB < — > Z5E, 

BC < — ► Iff, 

au ^ ra 



192 Congruence of Triangles Chapter 5 

We are now ready lo state the definition of a congruence between 
two triangles. 

Definition 5.1 Two triangles (not necessarily distinct) are 
congruent if and only if there exists a one-to-one correspond- 
ence between their vertices in which the corresponding parts 
are congruent. Such a one-to-one correspondence between 
the vertices of two congruent triangles is called a congruence. 

If ABC < — > DEF is a congruence between A ABC and ADEF, 
then we write A ABC at A DEF and note that the following six state- 
ments are true: 

I A aZD, AB^DE, 

in?- lE r BC=zEF, 

ACstLF, AC = DF. 

We can also say that, for A ABC and A DEF, if all of these six state- 
ments are true, then ABC * — * DEF is a congruence, or what means 
the same thing, AABC as ADEF, In view of the definition of con- 
gruent triangles, we sometimes say that "corresponding parts of con- 
gruent triangles arc congruent," 

If AABC == ADEF, explain why each of the following six equa- 
tions is true: 

m£A = m£D t AB = DE, 

m£B = m£E, BC = EF, 

tn£C=mlF t AC = DF. 

Note that if AABC and ADEF arc distinct triangles and if 
ABC < — > DEF is a congruence, then it is correct to write AABC ss 
ADEF t but that it is incorrect to write AABC = A DEE The state- 
ment "AABC ==; A DEF" is a short way of saying "ABC « — * DEF 
is a congruence;" it is a statement about a one-to-one correspondence 
between the vertices of two triangles. The statement "AABC = 
ADEF" is a statement that two sets arc equal; it means that AABC 
and ADEF are names for the same triangle. 

Note also that 

if AABC ss A DEF t then ABAC == A EDF, 

For AABC B ADEF means that ABC * — > DEF is a congruence; 
ABACS A EDF means that BAC < — * EDF is a congruence; and 
ABC 4 — > DEF and BAC < — > EDF are two ways of describing the 
same one-to-one correspondence between vertices. On the other hand, 
if ABC i — > DEFh a congruence but ABC < — > FED is not a con- 



5.2 Congruence of Triangles 193 

gruence between the two triangles, then it is incorrect to write 
AABCm A FED. 

In drawing figures it is convenient to label congruent angles and 
congruent sides of triangles with the same number of marks as shown 
in Figure 5-2. 
c 





Figure 5-2 

of the six congruences indicated in the figure are 

LA^LF, IB^IE, /.C^ID. 

Name the three pairs of congruent sides indicated in the figure. The six 
congruences tell us ABC < — > FED and A ABC m AFED. 

It is also helpful to label the side that Is opposite a given angle in a 
triangle with the same number of mark* used in labeling the angle, as 
has been done in Figure 5-2. For example, in AABC t L C and side^fB 
(which is opposite LC) both have the same number of marks. 



EXERCISES 5.2 



Exercises 1-6 refer to Figure 5-3. 





D Figure W 



EOF is not a, congruence. 



A V3 * * VI 

1. Is AABC ^ ADEF? Why? 

2. Explain why the correspondence ABC * — ► 

3. Is it correct to write AABC ss AEDF? 

4. Is it correct to write ABC A = AEFD? Why? 

5. Write four more Statements of congruence between the two triangles, 
each of which foikiws immediately from the fact that ABC * — » DEF 
is a congruence. 

6. Write three correspondences between the vertices of the two triangles 
such that each is not a congruence, 



194 Congruence of Triangles 



Chapter 5 



7. The following figure shows two scalene triangles with corresponding 
congruent sides and angles marked alike. 




Copy and complete the following correspondences so that the re- 
sulting statements are true, 

LKP * — * \T\ is a congruence, 
LPK « — > \J} is a congruence, 
KLP < — * [T] is a congruence. 
KPL 4 — » [TJ is a congruence. 
PKL * — > is a congruence, 
PLK <- — * [f] is a congruence. 

& In Exercise 7, why is the correspondence LKP * — * KST not a 
congruence? 

9. In the figure. AA£Ca APQB. Copy and complete the following 
statements by supplying the missing symbols. 




The correspondence A[T| * — * \]}QR is a congruence. 
Lh*k LP AE^^l 



10. In the figure, AABCs A DBG. List the 
six pairs of corresponding, congruent parts 
of these two triangles. 




5,2 Congruenco of Triangles 195 

1 1 . In Exercise 10, if £ ABC and Z DBC are distinct coplanar acute angles, 

does my B? bisect I ABD? Wiry? 

12. In Exercise 10 (with the figure appropriately modified), if L ABC and 
L DBC are coplanar obtuse angles, does ray BC bisect Z AHDP Why? 

13. The figure below shows eight triangles. If two triangles look congruent, 
assume that they really arc congruent. Write congruences between such 
congruent pairs of triangles. 





14. Without using a figure, list the six pairs of corresponding, congruent 
parts for the triangle congruence 

AEFK =s AABT. 

15. Without using a figure, list the six pair* of corresponding, congruent 
parts for the triangle congruence 

ARPSss ALSR 



196 



Congruence of Triangles Chapter 5 

16. For A ABC and A CAS, it is true that AH - $C, AC. - SA, #C as CA, 
Z A as IS, Z C as Z A, and Z A as Z G. Write a statement of con- 
gruence between these two triangles. 

17. If &DEFis a scalene triangle* prove that the statement 

ADEFsz ADFE 
is false, 

18. If A AST as ASIA, what special property does A A ST have? Draw a 
suitable figure for A AST". 

19. If A/Jfitf as A MEN, what special property does ALMA r have? Draw 
a suitable figure for ALMN, 

In Exercises 20 and 21, complete the proof of the following theorem. 

THEOREM Congruence for triangles is reflexive, symmetric, and 
transitive. 

Proof: Let A ABC be any triangle. Then IA as ZA» Z B £* Z A t and 
ZCas ZC by the reflexive property of congruence for angles. Also 

AA=?AB, WlrsBCt and AC as AC 

by the reflexive property of congruence for segments. Therefore A ABC as 
AAAC by the definition of congruent triangles. Therefore congruence for 
triangles is reflexive. 

20. Prove that congruence for triangles is symmetric 

21. Prove that congruence for triangles is transitive. 

22. In the figure at the right, 

CD 1 AB, 
AC as BC, 
AD = BD t 
LACD as Z BCD, 
m£A = SO, 
mZB= 00. 

Prove that A ACD as A BCD, 

23. In the figure at the right, 

Af is the midpoint of §7, 
Mi s the midpoint of 7ZF., 
KS as EJ> 
mZS=_r»ZJ, 
W J KE f 
IE ± KE. 

Prove that A SKM as A JEM. 




5.3 "If-Then" Statements 197 



5.3 "IF THEN" STATEMENTS AND THEJR CONVERSES 

Many of our definitions, postulates, and theorems have been stated 
in the "if-then" form. They have been statements of the type "If p, 
then tf," where p and q are statements, (Remember that a statement is 
a sentence which is either true or false.) In other instances we have 
used the phrase "if and only if ** in the statement of some of our defi- 
nitions, postulates, and theorems. 

As tin example of the use of the phrase "if and only if consider die 
definition of congruent segments given in Chapter 3: Two segments 
are congruent if and only if they have the same length. This statement 
is a conjunction of the following two statements: 

L Two segments are congruent if they have the same length. 

2. TVo segments arc congruent only if they have the same length. 

Statements 1 and 2 can be restated in the "if-then 1 * form as follows: 

3. It two segments have the same length, then they are congruent. 

4. If two segments are congruent, then they have the same length. 

We note two important things about these last two statements. 

(a) Each is written in the "if-then" form. The if-clause of each 
statement is the then-clause of the other. (This also means that 
the then-clause of each is the if-clause of the other.) 

(b) Both statements may be used in proofs. For example, if we 
know AB as n? and want to establish AB = CD, we can use 
statement 4 to justify our writing AB = CD, On the other hand, 
if we know AB = CD and wish to establish AB s CD t we can 
use statement 3 for justification. 

The two statements, J and 2, or alternatively 3 and 4, are called 
converses of each other. That is, the statement "If p, then q" is called 
die converse of the statement "If q, then p" and "If q, then p" is the 
converse of "If p.. then q" The converse of a statement in the "if-then" 
form can be obtained by interchanging the if- and then-clauses. 

If p r then (j. 

If g, then p. 
For example, the converse of the statement: 
"If 1 live in Seattle, then I live in the state of Washington" 
is the statement 

"If I live in the state of Washington, then I live in Seattle." 



198 Congruence of Triangles Chapter 5 

It is evident from this example that a statement and its converse need 
not both be true. 

When a definition is given in the "if-then" form, it is understood 
mat the statements of the definition and its converse are both true. As 
an example, consider Definition l.l of Chapter 1: 

Space is the set of all points. 

This definition can be restated in the "if-then" form as follows: 

1. If S is space, then S is the set of all points. 
The converse of (1) is 

2, If S is the set of all points, then S is space. 

Thus the complete definition is: If S is space, then S is the set of all 
points; and if S is the set of all points, then S is space. The "if and only 
if' form (which is logically equivalent to the complete definition) is; 
S is space if and only if S is the set of all points. 

Although a definition in * "if-then" form always implies the converse 
statement, this is certainly not true of all postulates and theorems. We 
discuss the "if-then" form of a theorem more fully in Section 5.4. 

As an example of a postulate whose converse is not true, consider 
Postulate 2 (The Line-Point Postulate). 

"Every line is a set of points and contains at least two points." 

We can restate this postulate in the "if-then" form as follows: 

"If I is a line, then / is a set of points and contains at least two 
points." 

The converse of this statement is 

"If I is a set of points and contains at least two points, then t is a 
line." 

Clearly, this last statement is false. There are many sets of points such 
as planes that contain at least two points and that arc not lines. 
Consider the following theorem proved in Chapter 4. 

"Vertical angles are congruent." 

In the "if-then" form, this theorem can be stated as follows: 

"If LA and L B are vertical angles, then LA === LB" 
Surely the converse of this theorem, staled as follows, is false: 

"If LA=* LB, then LA and / B are vertical angles." 



5.3 "If-Ttwn" Statements 199 

We conclude this section with some remarks about definitions and 
truth. A definition in our formal geometry is accepted as a true state- 
ment. Why is "Space is the set of all points" a true statement? It is true 
"by definition." Definitions help us communicate. It is helpful to have 
one word that means the same thing as "the set of all points/* It is help- 
ful to have one word to describe several points that all He on the same 
line. Why do eollinear points all lie on the same line? They do, by 
definition. 



EXERCISES 5.3 

In Fxerdses 1-5, a definition iu "if-then" form is given. Write its converse. 

L If the points of a set are eollinear. then there is a line which contains all 
of them. 

2. If there is a plane which contains all the points of a set, then those 
points are coplanar, 

3. If point A is between points B and C, then rays A3 and A C are opposite 
rays. 

4. If an angle is a right angle, then its measure is 90. 

5. If W is the midray of LABC. then BP is between BA and §C and 
ZAflFSB IFBC ' 

In Exercises 6-8, write the definition of the given phrase using the "if and 
only if form. 

6. Acute angle 

7. Linear pair 

8. Vertical angles 

In Exercises 9-16, is the statement true or false? Write the converse of the 
statement. Is the converse true or false? 

9. If two sets are convex, then their intersection is convex. 

10. If two angles are right angles, then they are congruent. 

11. If two angles are complements of congruent angles, then they are 
congruent, 

12. If 5» is the interior of an angle,, then S is a convex set. 

13. If two angles are congruent, then they are supplements of congruent 
angles. 

14. If I! st CD, then WstAB. 

15. If AABC^ AKLM, then AKLM St A ABC, 

lft If AABCss ADEF and AD£Fs ARST, then A ABC as ARST. 



200 Congruence of Triangtos Chapter 5 



5.4 THE USE OF CONDITIONAL STATEMENTS IN PROOFS 

As you have seen, many of the theorems have the form "tip, then 
</," where p and q are statements. Not all theorems are stated in this 
way, however, because it is sometimes easier to state them otherwise, 
but every theorem can be rostated in the "If p, then q" form. 

A statement of the form "If p> then tf is called a conditional The 
if-clause (the p statement) Is called the hypothesis and the then-clause 
(the q statement) is called the conclusion. In order to understand math- 
ematical proof better, we examine how such statements are used in 
proofs- 
Let us consider the following statements concerning two numbers 
x and y about which we know nothing except what we are told in the 
statement (A). 

(A) lix= y, then x 2 = y*. 

We see that (A) involves two statements. 

(B) x = y (the hypothesis) 

(C) x 2 = y 2 (the conclusion) 

Even though we do not know the replacements for x and y, can we say 
anything about the truth of statements (A), (B) s and (C)? Do we know 
that ( B) is true? Do we know that (G) is true? What about statement 
(A)? Your experience in working with numbers should help you to see 
that even though (B) need not be true and that (C) neeA not be true, 

(A) is true. Thus a conditional may be true even though its hypothesis 
and conclusion are not. Replace x and y with several pairs of numbers, 
some of which are equal and some of which are not. You should find 
that in those cases where you chose unequal numbers for x and y t both 

(B) and (C) are false and in 'those cases where you chose equal numbers 
for x and y, both (B) and (C) are true. But, in every case, (A) is true. 

Let us examine one such case where the replacements for x and y 
are unequal numbers. Suppose we replace x with 2 and y with 1 , Then 
statement (B) becomes 2 = 1 and statement (G) becomes 2 2 = I 2 , or 
4 = 1. Of course, both statements (B) and (C) are false. However, if 
we accept the hypothesis that 2 = 1, use the multiplication property 
of equality to write 3 • 2 = 3 • 1, or 6 = 3, and use the subtraction 
property of equality to write 6 -2 = 3- 2, or 4= 1, we have shown 
that the statement "If 2=1, then 2 2 = I 2 " is true. 

Of course, a general statement in the form of a conditional is of no 
value in a specific situation if the hypothesis of the conditional is false 



5.4 Conditional Statements in Proofs 201 

in that situation. The truth of "If p, then (f does not by itself guarantee 
tlie truth of either p or q. But the truth of the conditional and of the 
hypothesis is a different story, as we shaU see. 

Let us go back, then, to the three statements (A), (B), and (C) in- 
volving x and y. Suppose that (A) and (B) are both true. ITiat is, suppose 
that the conditional and the hypothesis are both true. Then it follows 
logically that the conclusion (C) is true. Check this with our example. 
This is a most important concept in mathematical proofs. It means that 
we can assert (C) after we have proved or know that both (A) and (B) 
have been established On the other hand, it does not mean that (B) 
follows from (A) and (C). (In our example, x = y does not follow from 
x 2 = ry 2 , since we could also have x = — y<) In general,, 

if a conditional and its hypothesis are both known 
to be true, then the conclusion of the conditional 
is also true. 

More concisely, if we know that the following two statements have 
been established; 

L if p. then q 

then we may conclude that q has been established. This means, in our 
example, that if we know the following two statements are both true: 

(A) if as = y, then x 2 = y 2 

(B) x=y 

then we can conclude x 2 = y 2 . 

How are you to know when a conditional is true? In our example, 
the conditional 

if x = y, then x 2 = y 2 

is a theorem that can be proved using the properties of numbers. In our 
geometry, a conditional is accepted as being true if it is a postulate, a 
previously proved theorem, or part of a definition. 
Let us look at two more examples. 

Example I We know that the conditional 

if JB m CD, then AB = CD 

is true. Why? Suppose that we also know or have been able to establish 
that AB ^ CD. What can we conclude? 



202 Congruence of Triangles Chapter 5 

Fjxamph 2 We know that the conditional 

if two angles are right angles, then they are congruent 

is true. Why? Suppose that you are able to establish that Z A and LB 
are right angles. What can you conclude? If, in connection with the 
same conditional, you are able to establish that LA^ LB, can you 
then conclude that £ A and Z B are right angles? Why? 



EXERCISES 5.4 

Write the theorems in Exercises 1-6 in "if-then" form. State the hypothesis 

and the conclusion of each. 

1, Supplements of congruent angles are congruent. 

2, Right angles are congruent 
H. Vertical angles are congruent. 

4. Two angles of a linear pair are supplementary. 

5. The intersection of two convex sets of points is a convex set. 

6. The interior of a triangle is a convex set. 

In Exercises 7-10, a statement p and a statement q are given. In each exer- 
cise, write the truth value {'that is, true or false) (a) for the statement "If p, 
then q" and (b) for the statement "If q t then p" The answers to Exercise 7 
have been given as a sample, 

% p: ml A = 90 and mLB = 90 

q; LA^LB 
(a) T (b) F 



8. p 

9. p 

9 
10. p 



VB is the midray of Z AVC. 

ZAVBsb LBVC 

A-M-B and AM -a MB^ 

M is the midpoint of 3B. 

Z A and Z B arc supplementary angles. 



(f LA and Z B are a linear pair of angles. 

In Exercises 11-2G, certain given statements are to be accepted as true. 
Then a conclusion is stated. In each exercise, state whether the conclusion 
is true, false, or inconclusive (that is, not enough information is given to 
decide whether the conclusion is true or false). 

11, Gkmi: x + y = 16, x - y = 12. 

Concision: x = 15, 

12, Given: If there is not u cloud in the sky, then it is not raining. 

There is not u cloud in the sky. 
Conclusion: It is not raining. 



5.5 Proof* in Two-Column Form 203 

13. Given: If iherc is not a cloud in the sky, then it is not raining. 

It is not raining. 
Conclusion: There is not a cloud in the sky. 

14. Given; Yon urc a member of the team only if you obey the training 

rules- 

Ken is a member of the team. 
Conclusion: Ken obeys the training rules. 

1 5. Given; You are a member of the teai 1 1 only if you obey the training 

rules. 

Bill obeys the training nilei. 
Conclusicm: Bill is a member of the team. 

16. Given: 2% + 3y = 12, x - y = 4 
Conclusion: 3x + 2u = 16 

17. Given: The intersection of two convex sets of points is a convex 

set. S and T are convex sets of points. S f\ T = H. 
Conclusion; R is a convex set. 

18. Given: If a and b are numbers and if ab = 0, then a = or b = 0. 

a and b arc numbers, ab = 0, and a =£ 0. 
Conclusion: h = 

19. (.'teen: If mLA = 30 and mZB s 00, then LA and Z B are 

complementary angles. 
ZA and LB are complementary angles. 
Conclusion: mL A = 30 and fnZB= 60 

20. Given: Linda will marry Joe only if he will buy her a new house. 

Joe will buy Linda a new house. 
Conclusion; Linda will marry Joe. 



5.5 PROOFS IN TWO-COLUMN FORM 

Students often ask, "What is a correct proof?" Unfortunately, there 
is no simple answer to the question. Making correct proofs is something 
that each of us learns by experience. A proof that may seem convincing 
to you may not be at all convincing to another person with much less 
experience in geometry than you. A deductive proof of a theorem is a 
set of statements, one or several or many, that shows how the conclu- 
sion follows logically from the hypothesis. To make a good proof it is 
important to think clearly about what is given and what is to be proved, 
and to consider various possibilities of statements which will lead from 
what is given to what is to be proved. It will help you considerably if 
you have a firm understanding of the postulates, definitions, and 
theorems already stated or proved. 

The proof of a theorem is often given in paragraph form. We have 
used this form of proof for most of the theorems proved thus far. For 
illustrative purposes look at the proof for Theorem 4.13 in Chapter 4. 



204 Congruence of Triangle* Chapter 5 

THEOREM 4.13 Any two right angles are congruent 

Proof: Every right angle has a measure of 90, Hence all right angles 
have the same measure, and hence they are congruent to each other. 

This proof consists of two sentences. Since the second sentence 
consists of two parts, there arc really three steps in the proof. These 
three steps, or links, form a chain of reasoning that shows how the 
conclusion, 

they are congruent, 

follows logically from the hypothesis, 

two angles are right angles. 

In writing this proof we did not provide reasons to support these three 
steps because we felt that the proof as given could be understood by 
someone with your background in geometry. If we were trying to con- 
vince someone with less practice, it would be necessary to give addi- 
tional statements for justifying each of the statements in the proof. 

One of the advantages of the paragraph type of proof is that it is 
not always necessary to give the reasons for all the statements when 
those reasons are obvious. However, when you give a proof in this 
form, you should be prepared to fill in the reasons. 

A two-column proof of a theorem consists of a chain of statements 
written in one column with a supporting reason for each statement 
written in a second column. When a statement in this chain is estab- 
lished because it is part of the hypothesis of the theorem, we simply 
write "hypothesis" or * 'given" as the reason. Otherwise, a statement 
may have as its supporting reason a combination of a conditional and 
its hypothesis. As stated before, a conditional is acceptable if it fa a 
postulate, a part of a definition, or a previously proved theorem. The 
hypothesis of this conditional should have appeared as an earlier state- 
ment in tile proof. The conclusion of the conditional should apply to 
the statement that is being supported 

A two-column proof of Theorem 4.13 follows. Note that in reasons 
which are conditionals, we write the numbers of the statements in 
which we have established the hypothesis of the conditional. 

THEOREM 4.13 Any two right angles are congruent. 

RESTATEMENT; If Z A and L B are any two right angles, then 

Proof: 

Hypothesis; £ A and L B arc right angles. 

Conclusion: £A ~ LB 



5.5 Proofs in Two-Column Form 



205 



1 . L A and Z B are right angles. 
2 mLA = 90, mlB a 90 

3. mZA = mZ£ 

4. IA^ LB 



1. Hypothesis 

2. If an angle is a right angle (1), 
then its measure is 90, 

3. Substitution properly of 
equality (2) 

4. If two angles have equal meas- 
ures (3), then they are con- 
gruent 



We note several important points about this proof. 

1. The proof is not complete until the last statement in the left- 
hand column is the same as the conclusion. 

2. When a statement Is part of the hypothesis, we write "hypoth- 
esis" or "given" as its reason, 

3. When a reason is in the "if-then" form, its hypothesis refers to 
an earlier statement or statements for support. For example, the 
if-clause of reason 2 refers to statement 1, However, the then- 
clause of reason 2 refers to statement 2. 

4. When a reason is not in die "if-then" form, and it can he written 
in that form, then it must satisfy the requirements stated in (3). 
For example, reason 3 simply states: "Substitution property of 
equality (2)," We could also have stated reason 3 as follows: "If 
a, b t and c are numbers and if a = c and b = c (2), then a = h." 

5. in proving the theorem we have not proved statement 4 con- 
sidered by itself as an isolated statement Rather we have proved 
the following conditional: If statement 2, then statement 4. 

Your teacher may permit you to list your reasons simply by identi- 
fying the postulate, definition, theorem, or property of equality which 
supports each statement. If this is the case, the proof of Theorem 4.13 
might read as follows: 



Statement 

1. L A and L B are right angles. 

2. mlA = 90, m£B- 90 

3. mLA = mlB 

4. £A==* IB 



1. Given 

2. Statement (1) and the defini- 
tion of right angle 

3. Statement (2) and the substi- 
tution property of equality 

4. Statement (3) and the defini- 
tion of congruent angles 



206 Congruence of Triangles 



Chapter 5 



We have shown three examples of proofs of the same theorem. 
Which is the best proof ? The answer is that they are all good. The two- 
column proof reminds us that we audi be able to give a reason for ever)' 
statement we make, and it also makes it easier to see which hypothesis 
we accept to begin our proof. Which proof you choose may depend on 
whom you are trying to convince. Usually in writing a proof your ob- 
jective will be to convince your teacher that you understand the proof. 
Your teacher may want you to have experience in writing both the par- 
agraph type of proof and the two-column type. 



EXERCISES 5.5 

1. In the following two-column proof several reasons are given in the 
"if-then" form. For each such reason indicate to which statement the 
if-clause refers and to which Statement the then-clause refers. 

THEOREM Supplements of congruent angles are congruent. 

restatement; If /. < r i s a supplement of Z a and if L. d is a sup- 
plement of Lb and if La == Lb, then Lc m L<L 



4- 



Proof; 

Hypothesis; L c is a supplement of L a, 

L d is a supplement of Lb. 

Lam Lh 

Conclusion: Lc ^ Ld 



dXb 



1. Z c is a supplement of L (t, 
Ld is a supplement of Z b, 

2. mLc+ mLa = ISO 
mLd+mLb = 180 

3. La== Lb 

4. mLa = mLb 

5. mLc + mLa = m/.d + 

fflZ b 

6. mLc = mLd 



7. Lczz Ld 



1. Hypothesis 

2. If two angles are supplementary, 
then the sum of their measures 
is 180. 

3. Hypothesis 

4. Tf two angles are congruent, then 
they have the same measure. 

5. If x, y, and z are numbers and if 
x = y and z = y, then x = z. 

6. If a, b, x, y are numbers and if 
a — b and x = «/, then x — a = 

y-b. 

7. If two angles have the same meas- 
ure, then they are congruent. 



5,5 Proofs in Two-Column Form 207 

2. Identity each of reasons 2, 4, 5 r 6, and 7 in Exercise 1 as a postulate, 
definition, theorem, or property of equality. 

3. Write a two-column proof for these theorems from Chapter 4, 
;a) Complements of congruent angles are congruent. 

In Vertical angles are congruent. 
4* Write a proof for the following theorem in (a) paragraph form and 
ib) two-column form. 

THEOREM If two angles are both congruent and supplemen- 
tary, then each is a right angle. 

5, Write a two-column proof of the theorem that vertical angles arc con- 
gruent using (a) the definition of vertical angles, (b) the definition of 
a linear pair, (c) the theorem that if two angles form a linear pair, then 
they are supplementary, and (d) the theorem that supplements of con- 
gruent angles are congruent (Hint: In the figure, let La and Z c he a 
pair of vertical angles. Prove Z a s Zc>) 




In Exercises 6-11, you are to perform some experiments with physical tri- 
angles. You will need a ruler, a protractor, and a compass. You are tv use 
these experiments as a basis for formulating the next three postulates in 
Section 5.6. Do the required constructions and measurements carefully and 
answer all the questions before proceeding to the next section. 

6. Construct a triangle, AAJ3C, in which AB = 2| in., AC = 1^ in,, and 
m/.A = 50, Measure the remaining three parts (BC, /_!&,/.€) of your 
constructed triangle and compare your measurements with those of 

two or three of your classmates. Are they nearly the same? 

7. Construct a triangle, ARST, in which HS = 2 in., m Z R - 40, and 
mZ S = (it). Measure the remaining three parts { Z T, RT t ST) of your 
constructed triangle and compare your measurements with those of 
some of your classmates. Are they nearly the same? 

8. Construct a triangle, APQR, in which EQ = 5 cm., FR = 4.5 cm., 
and RQ = 6 cm. (You may need to use a compass for this construc- 
tion.) Measure the three angles of your constructed triangle and 
compare your measurements with those of some of your classmates. 
Are they nearly the same? 

J). Construct a triangle, &DEF, in which mLD = 50, m£B = 60, and 

mZF = 70. Measure the three sides of your constructed triangle in 
centimeters. Compare your measurements with those of some of your 
classmates. Are they nearly the same? 



208 Congruence of Triangles 



Chapter 5 



10. Construct a triangle, &LMN, iii which m£L = 40, LM — 5 cm,, and 
M\ ! = 3.5 cm. (You may need to use your compass again for tills con- 
struction,) Measure Z M, L N, and LN of your constructed triangle and 

compare your measurements with those of some of your classmates. Are 
they nearly the same? 

11. Using only your ruler, construct a triangle, AADE, which has no two 
of its sides congruent. Now construct AA'DE* (distinct from AADE) 
such that AADE a A'D'F!. (You need not restrict yourself to using 
only your ruler for this construction.) How many of the six parts of 
AADE did you use in obtaining AA'D'E'? Could you have obtained 
AA'D'E by using a set of parts different from those that you actually 
used? What is the least number of congruent parts of AADE needed 
to be sure that AA'D'E 1 is congruent to AADE? 




5.6 THE CONGRUENCE POSTULATES FOR TRIANGLES 

We asked you to perform certain constructions in Exercises 5.5. 
We now wish to examine these con- 
structions in more detail, but first 
we need some definitions. In A ABC 
(Figure 5-4) we say that Z A is in- 
cluded by sides AC and "KB. Simi- 
larly, we say that side B€ is included 
by angles Z B and Z C 



Definition 5.2 An angle of a triangle is said to be included 
by two sides of that triangle if the angle contains those sides, 
A side of a triangle is said to be included by two angles of 
that triangle if the endpoints of the side are the vertices of 
those angles. 



_ln A ABC, shown in Figure 5-4, which angle is included by sides 
BC and BA? Which side is included by Z A and Z C? Were you able 
to answer these last two questions without looking at the picture of 
the triangle? Without looking at a pienre of ARST, state which angle 
is included by sides ST and RT, Which side is included by Z R and Z S? 
In Exercise 6 of Exercises 5.5, you should have concluded that all 
triangles having the given parts are congruent; similarly for Exercises 
7 and 8. When tins is true, we say that the three given parts determine 
a triangle. In Exercise 6, the three given parts of the triangle were two 
sides and the included angle. In Exercise 7, the three given parts were 
two angles and the included side, and in Exercise 8 the three given 
parts were the three sides. In Exercises 9 and 10 ? however, you should 
have found that not all triangles having the three given parts are con- 



5-6 Congruence Postulates for Triangles 209 



gruent. How many triangles of different size can be constructed if only 
the measures of the three angles are given (assuming there is at least 
one triangle with angles having these measures)? How many triangles 
of different sizes can be constructed using the data of Exercise JO? 

How did you construct AA'HE' in Exercise 1 1? Make a list of the 
steps you used. Perhaps you have one of the following lists. 



list I 

1. Draw AU^ 

2. Draw LA' = 
LA, 

3. DrawA'Fs 

4. Complete the 
construction by 
connecting R' and 
j r 7withD ? F, 



List2 

1. Draw ZA's 
LA 

2, Draw A' if ss 
AD. 

3, Draw LU zs, 
LD. 

4. Complete the 
constmction by 
drawing the sides 
ot LA' and LD 1 
long enough. 



List3 

1. DrawATFss 
AD. 

2. Draw an arc with 
A' as center and 
A'Ef as radius. 

3. Draw an arc with 
17 as center and 
D'E' as radius. 

4. Complete the 
construction by 
connecting the 
intersection of the 
arcs to A* and U. 






A' D' a* n- A' d- 

For each list, the figure at the bottom shows what the construction 
looks like just before it is completed. 

Tjook at the first list. How many side measures are used? How 
many angle measures? Is the angle between the two sides? This com- 
bination oitwo sides and the included angle is abbreviated by the sym- 
bol S.A.S.; the correspondence ADE + — ► A'UE' is referred to as an 
S.A.S, congruence because we feel that this much information about 
congruent pairs is enough to guarantee that all matched pairs of parts 
are congruent. We make this conclusion formal in Postulate 23, 

Look at the second list. What combination of measures is used? 
This combination of two angles and the Included side is abbreviated 
A.S.A.; the correspondence ADE* — >A'D'E' is referred to as an 
A.S.A. congruence. We make this conclusion formal in Postulate 24, 

Look at the third list. This combination of three sides is abbreviated 
S.S.S., the correspondence ADE< — *A'D'E' is referred to as an 
S.S.S. congruence. We make this conclusion formal in Postulate 25. 



210 Congruence of Triangtes 



Chapter 5 



POSTULATE 23 (The S.A.S. Postulate) Let a one-to-one cor- 
respondence between the vertices of two triangles (not necessarily dis- 
tinct) be given. If two sides and the included angle of the first triangle 
are congruent, respectively, to the corresponding parts of the second 
triangle, then the correspondence is a congruence. (Sec Figure 5-5.) 





A ABC 



fl D 

ADEF hv the S.A.S. Postulate 



Figure 5-5 



POSTULATE 24 (Tkts A.S.A. Postulate) Let a one-to-one cor- 
respondence between the vertices of two triangles (not necessarily dis- 
tinct) be given. If two angles and the included side of the first triangle 
are congruent, respectively, to the corresponding parts of the second 
triangle, then the correspondence is a congruence. (Sec Figure 5-6.) 





A ABC = ADEF by the A.S.A, Postulate 

Figure 5-6 

POSTULATE 25 (The EMS Postulate) Let a one-to-one cor- 
respondence between the vertices of two triangles (not necessarily dis- 
tinct) be given. If the three sides of the first triangle are congruent, 
respectively, to the corresponding sides of the second triangle, then 
the correspondence is a congruence. (See Figure 5-7.) 





B D 

A ABC * ADEF by the &&& Populate 



Figure 5-7 



5,6 Congruence Postulate* for Triingles 211 

Note that there is no A.A-A. Postulate (Figure 5-8) and no S.S.A. 
Postulate (Figure 5-9). Also note how the results of Exercises 9 and 10 
of Exercises 5.5 arc related to this statement. 




B D 

AABCgt ADEF 






AAHC £ ADEF 



Figure 5-9 



Our experience with physical triangles suggests that it would be 
proper to include an S. A. A. Postulate (Figure 5-10). Actually, we do not 
need such a postulate. The statement that you might expect as a postu- 
late is, in fact., given as a theorem in a later chapter. We defer the proof 
because we need not only the congruence postulates but also a postu- 
late about parallel lines, which appears later, before we can prove the 
S.A.A. statement. Since it is easy to prove later, we have decided not 
to adopt it formally as a postulate. 





AAiJC := AJ>EF b> S.A.A. 



Figure 5-10 



The A.S.A. and S.S.S, Postulates can be proved as theorems once 
the S.A.S. Postulate is assumed. The proofs are difficult, however, and 
so we have adopted these statements as postulates in order to make 
simpler the development of our geometry. 



212 Congruence of Triangles 



Chapter 5 



EXERCISES 5.6 

In Exercises 1-16, like markings on the triangles indicate congruent parts. 
In each exercise, determine if a pair of triangles can be proved congruent. 
If a congruence can be proved, write a triangle congruence (in the form 
ABAC ss &FDE) and the postulate (S.A.S., A.S.A., S.S.S.) you would use 
to prove it. Exercise 1 has been worked as a sample. 



J. 




ABAC at A FDE 
by&A£. 




h D 




3. 




i *i» 





6. R 





7. 




5.6 Congruence Postulate! for Triangles 213 
13, V 




214 Congruence of Triangles 

17. For A ABC and ADEF, 



Chapter 5 



■\B - EF, 
I As:- IE, 
AC^ED. 

Write a congruence between the two triangles. What postulate are you 

using? 

IS. For ARSUmd &GKL, 

Which two sides must be proved congruent if 

AJRl/5 == &LKG 

by the A.S-A, Postulate? 

19. For AFQR aud AQSR, 

FRsiSn and PQ^SQ, 

Arc the triangles necessarily congruent? Why? Draw a figure, 

20. For A ABC and A BCD, 

IABC=* IDBC and LACB== LDCB. 
Are the triangles necessarily congruent? Why? Draw a figure. 

In Exercises 21-30, two triangles appear to be congruent in the given figure. 
Iu cadi exercise, certain information is given about the figure. Assume that 
all points are coplanar and have the relative positions shown. 

(a) Copy and mark each figure, as was done in Exercises 1 - 1 (i, to show 
the given information. 

(b) If the given information is sufficient to prove the triangles con- 
gruent, state a congruence between the triangles and the postulate 
you would use (S. A.S., A.S.A., S.S.S.) to prove it. If the given infor- 
mation is insufficient, write "Insufficient." 

£1. M is thcmidpoinl of FQ. 22. A~D - EC at D. 

PR==QR D is the midpoint of EC. 





5.6 Congruence Postulate? for Triangles 215 



23. VA is the midray of IBVC. 
m£VAC=mlVAB 




27. L-F-W N-E-M 

D is the midpoint of LM, 
LF = ME 
DF = DE 

N 



24. Consider onlv APQR and ASRQ. 
P0 _L UK, SR 1 OT, and 




25. Consider onlv ARTS and AR VS. 
KT^RV 
ST=sW 

m, STV = ml$VT 
mlRTV = mARVT 





L D M 

In Exercise 27, if, in addition 
to the information given for 
the figure, we also haveLA T = 
MN, is AFNDatAEXD? 
Why? (Hint: Use ND in your 
proof. This segment exists 
even though it is not shown in 
the figure.) 

N-H-M 

U is the midpoint of GK. 
IsG LGK 
MR 1 GK 

N 



26. A-E-D, D-F-C 

B is the midpoint of AC. 
BE 1 AD 
RF 1 £ft 
AE = CF 




30, TR ± 

VR ^ SR 
A-R-B 





216 Congruence of Triangles Chapter 5 

5 J USING THE S,A.S. r A.S.A., AND S.S.S. POSTULATES 
IN WRITING PROOFS 

In this section and lire following one, you will be asked to write 
your own proofs. In writing these proofs you will usually need to prove 
one or more pairs of triangles congruent by using the S.A.S., A,S,A.„ 
and S.S.S. Postulates. Therefore, in planning your proof, you should 
look for the opportunity to apply one of these postulates to some pair 
of triangles. We illustrate with some examples. 

Example 1 If M is the midpoint of AB and CD, then AC ^ ED. 

In starling to construct a proof for this statement, we first draw a 
figure which seems to fit the hypothesis. 
c 




Figure Ml q 

Figure 5-11 shows M to he the midpoint of segments A~R and CD, 
(Can you draw a different figure which shows the same information?) 
We have marked AM and JIB with the same number of marks since, by 
the definition of a midpoint, we know thai A^f s MB. CM and MD 
have been marked alike for the same reason. 

Before attempting to construct a proof of a theorem, it is helpful to 
have a definite plan in mind. In this example, we want to prove that 
two segments are congruent. We know that two segments are congru- 
ent if they are corresponding sides of congruent triangles. Our plan is 
to prove A AMC = ABMD in Figure 5-11. For completeness the fig- 
ure is included as a part of the proof, as it should be. 

Pmof: 

Hypothesis: M is the midpoint of AB and CD. 

Conclusion: AC = H75 

(Plan; Prove AAMC £s &BMD.) 




5.7 Using the S.A.S., A.S.A., and S.S.S. Postulates 217 



Reason 



1. M is the midpoint of AB and 

CD. 
2.ZMe*m. CM a W) 
3. £AMC=z I BUD 
4 AAMCj^ ABAID 
5. AC ^ tf D 



L Hypothesis 

2. Definition of midpoint (1) 

3. Vertical angles arc congruent 

4. S.A.S. Postulate (2. 3) 

5. If two triangles arc congruent 
(4), then their corresponding 
parts are congruent 



Note that there are essentially the following five steps to writing a 
geometric proof in two-column form. 

L Draw a figure which seems to fit the hypothesis and, where pos- 
sible, mark on the figure the information given in the hypothesis. 

2. State what is given (the hypothesis) expressed in terms of the 
figure. 

3* State what is to t>c proved (the conclusion) also expressed in 
terms of the figure, 

4, State a plan for the proof. The plan need not he expressed in 
written form, but should be carefully thought through before at- 
tempting to write the proof. 

5. Write out the proof in two-column form, 

Now let us examine the proof for Example 1. You probably noticed 
in the statement of Step 3 that we relied strongly on the figure for giv- 
ing us the information that L AMC and Z BMD are a pair of vertical 
angles. If our figures are carefully drawn, we can rely on them to give 
us correct information, However, we must always be prepared to he 
able to justify any information that we take from a figure by postulates, 
definitions, and theorems. In the figure for the example, we know that 
points A, Af, B are eollincar and that points C, M, D are collmear 
(Why?). We also know that when two lines intersect, vertical angles 
are formed. Since the figure clearly shows all of this information, we 
did not bother to establish in our proof that I. AMC and Z BMD arc 
vertical angles. 

Since this may be your first experience in writing geometric proofs 
in two-column form, your teacher may want you to write a more com- 
plete proof than the one given. In other words, your teacher may not 
want you to assume any information from a figure without establishing 
this information in your proof. If this is the case, we give a second, more 
complete proof of the theorem in Example 1 for your consideration. 



2 IS Congruence of Triangles 

Ifajjxithesis: M is the midpoint of AH and CD. 
Conclusion: AC =k BD 



Chapter5 




1, M is the midpoint of A73 and 
CD. 

2, M is between A and B, and M 
is between C and D. 

3, Ml, M& and m£> m£> are two 
pairs of opposite rays. 

4, IAMC and ZIttfD are ver- 
tical angles. 

5, LAMCm IBMD 

e.A~M==MB,CM=*MD 

7. AAMCss AJ3MD 

8. A~€=±BD 



1. Hypothesis 

2. Definition of midpoint (1) 

3. Definition of opposite rays (2) 

'1 Definition of vertical angles 
(3) 

5. If two angles are vertical (4), 
then they arc eongnient. 

6. Definition of midpoint (1) 

7. S.A.S. Postulate (5, 6) 

8. If two triangles are congruent 
(7), then their corresponding 
parts arc congruent, 



Note that some of the reasons are written in abbreviated form. For 
example, reason 4 is ''Definition of vertical angles'* rather than the com- 
plete statement of this definition. Be quite certain that you know the 
complete statement before using the abbreviated form in your proofs. 

Example 2 % in quadrilateral A#CD, TLB m BC and AB a CD, then 

la = ia 

Hypothesis: AD a 1$C 

31 as ED 

Conclusion: LA^ LC 

(Plan: We want to prove L A = L C by showing they are corre- 

l ■ nding angles of congruent triangles. But there are no triangles in our 

figure. We therefore draw DB to show which triangles wc shall use/) 




S.7 Using the S.A.S.. ASA, and S.S.S. Postulates 219 



Note that the segment JDH in Figure 5-12 is dashed to distinguish 
it from the parts of the figure given in the hypothesis. 




Figure 5-1S 

We call a segment such as DB an auxiliary segment. Thus an aux- 
iliary segment is a segment that is not a part of the figure given in the 
hypothesis, but does exist by the definition of a segment and the Point- 
Line Postulate. Such segments should be drawn into your figure when 
it is convenient to use them in the proof. We now continue with our 
plan. 

Since two sides of AABD are congruent to the corresponding sides 
of ACDB, we would expect to prove these triangles congruent by ei- 
ther the S.A.S, Postulate or the S.S.S, Postulate. If we were to use 
S.A.S., we would need ZA^ L C. But since this is what we are trying 
to prove, we cannot use it as part of our proof- That leaves the S.S.S. 
Postulate. In AABD, the third side is BD and in A CDB t the third side 
is DB. But BZ5 and DB are the same segment and are congruent by the 
reflexive property of congruence. We can now write the proof. 



1. 35 s CD 

2. AD ^ CB 

3. Segment BD exists. 

4 ED~.T5B 

5. AABD =* A CDS 
a lAs IC 



1. Hypothesis 

2. Hypothesis 

3. Point-Line Postdate and the 
definition of segment 

4. The reflexive property of con- 
gruence for segments. 

5. $.$£, Postulate (I. 2, 4) 

6. If two triangles are congruent 
(5), then their corresponding 
parts are congruent. 



Note that when we wrote AD ^CSin statement 2 and BD s DB 
in statement 4, we were adhering to the correspondence ABD < — * 
CDB for the two triangles. Of course, it would be correct to write 
315 m EC In statement 2 and BD m BD in statement 4. But it helps 
to keep tilings clear for both the writer of the proof and someone read- 
ing it if it is written the way we wrote it, keeping the order of the letters 
consistent with their order in the congruence we wish to prove. 



220 Congruence of Triangles 



Chapter 5 



Example 3 If points are as shown in Figure 5-13 and if AB & CB f 
FE ?~ DF, and ~BE ^ ED. prove AF = FC. 

How do we formulate a plan for 
this proof? A plan for a proof can 
often be formed by "working back- 
wards." That is, we begin with the 
conclusion and Lry to work our way 
back to the hypothesis, If these 
steps can then be reversed, we have 
a plan for our proof. This method 




LonsisEs ui writing ^or cnizuung 
information in two columns: 


or; A c 

Figure 3-13 


I Can Prove 


If 1 Can Prove 


1. JTszFC 


L (a) AE^CD? 




(b) FEzzUF\/ 


2. JEseCO 


2, AAB£ss ACBD? 


a AABE=*ACBD 


3, (a) AB s CB (S))/ 




(b) ZB^Z/?(A)v/ 




(c) £EsAD(S)V 



read line 1 as follows: I can prove AF ^ FC if I can prove 
(a) AE at UD and (b) FE ^ DF. (Note that if we know statements (a) 
and iTj) of line 1 in the second column, then we ran deduce statement. I 
in the first column from Corollary 3.4.3.) Because (b) is given in the 
hypothesis, it is checked. Statement (a) is then to be considered. It is 
brought down to line 2. We read line 2 as follows: "1 can prove AE = 
<OD if 1 can prove A ABC ss ACBD" Since this triangle congruence 
is not given, we bring it down to line 3. Line 3 is read as follows: l T can 
prove A ABE ^ ACBD if I can prove (a) AB 3 CB, (b) IB == LB, 
and (c) HK "^ ED." Since all three of these statements are given or can 
easily be proved, they are checked. We can now write the proof by re- 
le order of the statements. 



Proof: 

Hypothesis: AH ^ CB 
FE^UF 
BE^ED 

Conclusion: AF ss FC 




Statement 



1. BEsBD 

2, ZBs IB 

4. AABE = ACUD 

5.AEaCD 



6. FE a DF 

7, AFaJU 



5.7 UsingtKe S.A.S., ASA, and S.S.S. Postulate* 221 
Reason 

1. Hypothesis 

2. Reflexive property of congru- 
ence for angles 

3. Hypothesis 

4. S.A.S. Postulate (1, 2, 3) 

5. If two triangles arc congruent 
(4), then their corresponding 
parts are congruent, 

6. Hypothesis 

7. Corollaiy 3,43 (5, 6) 



Example 4 In Figure 5-14, all points are coplanar. 
Hypothesis: 

C is in the interior of L ADE. 
E is in the interior of L CDB. 

W IDE 

W is CD 




Copy and complete the proof 
that AE=zC!B. 

{Plan: We plan to use ADE 
/.ADEel LCDB by the Angle Measure Addition Postulate, and 
DE 3: UB. We can then use SAS.) 



Figure 5-14 
-* CDB. We can show ID == CD, 



L KD _L DC, BD X DE 1. Hypothesis 

2. ml ADC = 90, mlBDE = 90 2. [?] (1) 

3. C is in the interior of Z ADE. 3. [?] 
E is in the interior of L CDB. 

4. ml ADE = 90 + mlCDE 
mlCDB = 9() + mlCDE 

5. mZAD£ = mZCDB 



6. ZADEj^ZCDB 

7. AD s= CD, DE s Dft 

8. AAEEtt A CDB 

9. 32 fi CB 



4, Angle Measure Addition Pos- 
tulate (2, 3) 



5, Substitution 
equality (4) 

& mo) 

7* 

8. {6, 7) 

9. eke), m 



property of 



222 Congruence of Triangles 



Chapter 5 



EXERCISES 5.7 



In Exercises 1-22, write two-column proofs. In each exercise, copy the fig- 
ure and mark on it the given congruences. 

L In the figure, S is between P and Q, BS ^ PQ, and Zla L 2. Prove 
that S is the midpoint of FQ. 




Copy the following outline and supply the missing reasons, includ- 
ing the numbers of supporting statements. 



Statement 



Reason 



1. ll^JJl 

2, BS=RS 
a RS ± PQ 

4. jLHSPandARSQ are right 
angles. 

5. /.RSF = £RSQ 

a APSil^ &QSB 

7. PS == P 

8. S is between F and Q. 

9. S is the midpoint of P%). 

2. In the figure, LARD and Z CBE 
are vertic al an gles, £ is the mid- 
point of DE, and ID St LE. 
Prove that B is the midpoint of 



3. In the figure, point B is between 
points A and C and point C is 
between points B and D. Given 
CJR ^ DS, Zls Z 2, and 
AB m CD, copy and complete 
the proof that Zr =* IS. 



1. Hypothesis 

3. S 

4. If lines determined by two seg- 
ments arc perpendicular (pi), 

then [7]. 

5. Any two right angles are [?] 

7. If {[UK then [g. 

am 

9. HKSE) 




5.7 Usng the S.A.S., A.S.A.. and S.S.S. Postulate* 223 



1. B is between A and C, and 
C is between B and D, 

2. KB s CD 
a EC*BC 

7. AACrt a ABDS 



1. Hypothesis 

^ m 

a ru 

4. Length-Addition Theorem for 

Segments (1, 2, 3) 

''• 'ill 

6. Hypothesis 

8, OiiTCsponding parts of congru- 
ent triangles are congruent 

(0D- 



4. In the figure, A is between C and D, B is be- 
tween C and K, K is (he midpoint of AB. 
L\ = Z 2, and AC = BC. Copy and com- 
plete the proof that LACK =z ABCK, 




L Zl a Z2 

2. D-A-C and E-B-C 

3. /, 1 and Z G&K form a lin- 
ear pair, and Z2 and 
Z CBK form a linear pair. 

4. Z CAIC is a supplement of 
Z I, and Z CBK is a supple- 
ment of Z2. 

5. ZCAKc-i ZCBK 

0. K is the midpoint of XB. 
7. [?]*»[?] 

9. ttlsiBC 

10. ACAK a AGBK 

11. ZACK a ZfiCK 



1.0 

2, Hypothesis 

3, Definition of linear pair (2) 



4. If two angles form a linear pair 
(3), then they are supplemen- 
tary. 

5, The supplements of congruent 
angles are \J]ih 4)« 

6. rj] 

7. Definition of midpoint (6) 

8. Hypothesis 

9. Definition of congruent seg- 
ments (8) 

io, rr|{[?],[?j,rT|) 
ILEKB) 



Chapter 5 



224 Congruence of Triangtes 

5. In the figure, LR = RN and 
LM = M-W Prove that 
mZL== m£N. 



A. Hypothesis: For A ABC, 5? is the midray of L ABC intersecting AT: 

in D. AC is the midray of I BAC intersecting EC in E. Points R, A, R r S 
are collinear in tliat order. L HAD m ZSB& 





Conclusion: A~E=*W) 

(Flan: Show that IS ss IS, that ZABE== /BAD (supplements of 
congruent angles); and that vilBAE = mlABD (halves of equals), 
Then use the AAA. Postulate to prove AABE B A BAD.) 

7. GStant DA = CB 

CBXA^ 
To Prove: /L&m£C 

8. Docs your proof for Exercise 7 de- 
pend on points A, B, C, and D being 
coplanar? 

ft Given: £lm Z2 
Z3 = Z4 
To Proce; PS = RQ and 

mZS= mZQ p 



10. c liven: D and K are between 
A and B as shown hi 

the figure, 

zage« zjbcd 
dc=?EC 

lb Prone: AAC& =g A BCD 
AEsSD 



1 1. Use the same hypothesis and figure given in Exercise 10 and prove 
(a) AACD m ABCE. (b) AS a TIE. 




12, In the figure, E is in the interior of 
Z BA C, A-D-E, Z EDO' cs Z EDB, 
and BD s CU Prove that A~S is 
Lhemidrayof ZRAC 



5.7 Using tht S.A.S., A.S.A,, and S.S.S, Postulates 

13. Given; 



us 




AB = CB 
AD = CD 
To Prove: l\= £2 
[Hint: DrawBD.) 




14. In the figure, R, S, 7', and N 
are collinear in that order. 
£BSV=± ZFJTvMSs W t 
TTa VS. 

From L?=LV 




s 




N 



15. In the figure, A-E-C, D-E-B, 
ZI» Z2, Z3ss Z4,and 

Prove that £C and £15 bisect 
each other at E. 





16. In the figure, E is the midpoint 
of J5F, H is in the interior of 

Z DEC, G is in the interior of 

Z FEW. Z DEG s Z FEW, and 
TJEssGE. Prove that Z tf =* Z G- 

17* Given the situation of Exercise 16, prove in two different ways that 
LDEIls* LFEG. 



18. Given: Zl=* Z2 

Af is the midpoint of R-S 

rt^st 

Prove: Z3=*Z4 




£26 Congruence of Triangles 



Chapter 5 



19. In thefigure^AD = BC, E is the midpoint of CD. and AE s BE, Prove 
that AC m& BD. 




A B 

Copy and complete the statements in the fallowing plan. 

I Can Prove If I Can Prove 



1, AC s£ BD 

2. A ADC's A BCD 



1. AADCssS A BCD? 

2. (a) [U-[2]{S)v' 

(b) ^=m{A)? 

(c) DC=CD{S)>J 

3. AEDAs A£CB? 
4 (a) ££^[7] 

( C ) [U-[i](S)v 

\ T ovv reverse the steps in the second column and complete the proof. 



a LADC s /. BCD 
4. A[?J & A[7] 



20. In quadrilateral A BCD, M is the midpoint of AC and Z 1 Si Z2, Prove 
that any segment which contains M 
and has its endpoints in the inte- 
riors of CD and AB is bisected at M, 
[Hint: Draw EF such that E-M-F, 
A-E-B, and D-F-C. Now prove 
that AEMAmt AFMC and hence 
that EM = FA-/ . i 




21. CHAUJ^CE PROBLEM. 1 1 1 A ABC T AC SS ATI 

Prove that the correspondence CAB *. — * 
BACisa congruence using the S.A.S. Postulate. 
Hence conclude that A CAB ^-- ABAC Does 
tills prove that IBsz Z C? Why? Does this 
prove that if a triangle has two congruent 
sides, then the angles opposite those sides are 
congruent? 




5.8 Isosceles Triangle* 227 



22. CHALLEN'CF PROBLEM. Ill AH ST, ZSsZF. 

Prove thul I he correspondence 5Tft « — * TSft 
is a congruence using the A.S.A. Postulate. 
Hence conclude _that ASTR m ATSfl, Does 
this prove that R$ =s JIT? Why? Does ihis 
prove that if a triangle has two congruent 
angles, then the sides opposite those angles are 
congruent? 




5.8 ISOSCELES TRIANGLES 

Each triangle shown in Figure 5-15 has at least two congruent sides. 
Such triangles are called isosceles (from the Greek words iso and skelm 
meaning "equivalent" and " legs," respectively'!. 





Definition 5.3 An isosceles triangle is a triangle with (at 
least) two congruent sides. If two sides are congruent, then 
the remaining side is called the base. The angle opposite the 
base is called the vertex angle. The two angles that are op- 
posite the congruent sides are called the base angles. 



In A ABC of Figure 5-15, A B and ~KC are congruent sides. The base 
is JJQ the vertex angle is Z A. and the base angles are Z B and Z C. 
Name the base, vertex angle t and base angles of ADEF in Figure 5-15, 

If a triangle has three congruent sides as does &PQR in Figure 
5-15, then any side may be considered as a base of the t dangle. The 
angle opposite a base is considered the vertex angle corresponding to 
that base, and the angles that include a base are called the base angles 
corresponding to that base. 



228 Congruence of Triangles 



Chapw 5 



Definition 5.4 A triangle with three congruent sides is 
called an equilateral triangle. A triangle with three congruent 
angles is called an equiangular triangle. 



If you worked Exercises 21 and 22 of Exercises 5.7, you already 
proved the following two theorems. 

THEOREM 5.1 (The Isosceles Triangle Theorem) The base an- 
gles of an isosceles triangle are congruent 

restatement: In A ABC (Figure 5-16), AC Si A75. 
Prove that ZDs LC. 



Flgur* 5-16 U 




(Plan: We will show that, for AABC, the correspondence 
BAC * — ► CAB is a congruence using S.A.S., and hence LB = LC 
by the definition of congruent triangles,) 

Proof: 



Statement 

1. XCzzAB 

2. ZAa LA 

3. tt^XC 

4. &BACm ACAB 

5. LB^ LC 



1. Hypothesis 

2. Reflexive property of congru- 
ence for angles 

3. Symmetric property of con- 
gruence for segments (1) 

4. S.A,S. Postulate (1.2, 3) 

5. Corresponding parts of con- 
gruent triangles (4) are con- 
gruent. 



Theorem 5.1 implies the result which follows. Its proof is left as 
an exercise. 



5.8 Isoitttas Triangles 229 

COROLLARY 5.1J If a triangle is equilateral, then it is 

equiangular. 

We know from Theorem 5.1 that if a triangle has a pair of congru- 
ent sides, then the angles opposite these sides are congruent. The eon- 
verse of Theorem 5.1 is also true, and we state it as our next theorem. 

THEOREM 5.2 (Converse of the Isosceles Triangle Theorem) 
If a triangle has two congruent angles, then the sides opposite these 
angles are congruent and the triangle is isosceles. 

bestatkmevt: In A DEF (Figure 5-17), LE^ LF, Prove that 
DF = DE and hence that A DEF is isosceles, 




F Figures 5-17 

(Plan: We will show that, for A DEF, the correspondence 
EFD < — ► FED is a congruence using A.S .A. and hence DF » DE 
by the definition of congruent triangles,) 

Fnwf: 



1. ZEss IF 

2. EF^TE 

3. ZF» IE 

4. AEFD BS AFED 

5. DF=sDE 



6. ADEF is isosceles, 



1. Hypothesis 

2. Reflexive property of congru- 
ence for segments 

3. Symmetric property of con- 
gruence for angles (1) 

4. A.S.A. Postulate {1, 2 f 3) 

5* Corresponding parts of con- 
gruent triangles (4) are con- 
gruent. 

6, Definition of isosceles triangle 
(5) 



The proof of the following corollary of Theorem 5.2 Is left as an 
exercise. 

COROLLARY 5.2.1 if a triangle is equiangular, then it is 
eqi 



2 30 Congruence of Triangles 



Chapter 5 



EXERCISES 5.H 

h Is every equilateral triangle isosceles? Is every isosceles triangle 
equilateral? 



2. In the fipnjre. F is in th<-' interior of 
A ABC, AB =* AC and FB ss PC 
Prove, without using congruent 
triangles, that ZABPftZACP. 




3. Write a two-column proof of Corollary 5.1,1. 

4. Write a proof, in paragraph form, of Corollary' 5.2,1, 

5. Prove Theorem 5.1 using the S.S.S, Postulate. 

6. In the figure, A RST is an isosceles 
triangle with vertex angle at H. 
Give a proof different from that 
given for Theorem 5.1 that base 
angles of an isosceles triangle are 
congruent (Hint: Let M be the 
midpoint of ST and prove 



7. Given: F-Q-S, F-R-T, and 
£IS3 Z2 

l^ve: &QPR is isosceles, 




8. Civen: A-B-D t A-C-E, 
AH =s AT?, and 

Pnwe: AflDCsACES 





9, challenge problem. Did you use the Isosceles Triangle Theorem in 
your proof of Exercise 8? If not, write a different proof, using the Isos- 
celes Triangle Theorem* If you did, write a different proof in which the 
Isosceles Triangle Theorem is not used. 



10. Given; A/)£F is isosceles with 
vertex angle at D, 

Prove: (ft) AGDH is isosceles, 
(b) Z UGH =s I DUG 



5.8 lio&celes Triangle* 231 
D 




K G 



H F 



1 1. If points A, B, C, A F, F have the 
betweenness relations shown in 
the figure, if A ABC is isosceles 
with vertex angle at li, and if 
A AFC £s isosceles with vertex an- 
gle at F* prove AD as EC (Phn: 
Use the Isosceles Triangle The- 
orem to prove A A DC fe ACEA 
by A.S.A.) 




12. In convex quadrilateral ABCD, 
JS^AT) and 55 a; W< (a) 
Prove I ABC ^ I ADC widiout 
proving any triangles congruent, 
(bj Draw AC intersecting B~D at 
E. Prove BE = Wand AC ± ED. 




13w In the figure, Z 1 ^ Z 2, points B, I>, F„ C are collinear. 
Prove that AABC is isosceles. 



BD = EC 




232 Congruence of Triangles 

14. Given: ARST is equilateral. 

Prove: ARST is equiangular. (Plan: 
Show that the correspond- 
ence ilST « — » TRS is a 
congruence by S.S.S, and 
that IR=*IT&£S by 
the definition of congruent 
triangles.) 



t5. Given: APQR is equilateral, with 
A, R. and C the midpoints of 
i 3 ^, ^H, and M, respec- 
tively. 
Prove: AARC is equiangular. 



Chapter 5 




1 0, challkn Gg l'noiiLEM. We can prove the A . S. A. Post ill ate as a theorem 
once we have assumed the S.A.S. Postulate. Complete the proof of the 
following statement. 

Given an A.S.A. correspondence 

FAM < — » RSP 

as indicated in the figures, if 

ZFs Zfl, 

IFAM== ZS, 
prove that A FAX! ^ A RSP. 





Proof: There is a point If on FM such that FM' ~~. HP. Why? (Note 
that our figure shows ,\f and Sf to be different points. We will prove- 
that they are the same point.) Therefore AFAM* ss. A RSP by S.A.S- 

{Show this.) IFAM' S I RSP. Why? Therefore AA$' = A*f by the 
Angle Construction Theorem, It follows that M f = M (Why?) and 
AFAM m ARSF. 



5,8 lsosc«le* Triangtes 233 



17, ciiam-Ekce problem. We can prove the S.S.S. Postulate as a theorem 
once we have assumed the 5.A.S. Postulate and proved the Isosceles 
Triangle Theorem. Complete the proof of the following statement. 
Given an SS.S. correspondence 

ABC <-h> PQH 

as indicated in the figure, if 



AC =r VH, 



prove that A ABC a APQR. 




1. AB = FQ t B~€~i §R r 

Wmm 

2. There is a point F on the 

opposite side of An from C 
such that ABAFr- £P> 

3. There is a po int D on AF 
such thatAD^ PR. 

t &ABD- s\PQR 



1. Given 

2. Angle Construction Theorem 



3. Segment Construction The- 
orem 

4. 5.AS Postulate (1,2, 3) 



Since C and D are on opposite sides of AB, CD intersects AB in a 
point H. Our figure shows H to he between A and B. We could have, 
however, A = il.avB = H, or A -B U, or H-A-B, Complete the proof 
for the case where A-H-B by using the Isosceles Triangle Theorem and 
the S.A.S. Postulate to prove that A ABC st AABD, It will then follow 
from step 4 and the transitive property of congruence for triangles thai 
A ABC" = &.PQR* Draw a figure and complete the proof for the other 

cases, that is, for A = 1L B = H, A-B-H, and II-A-B, 



234 Congruent* of Triangles 



Chapter 5 



5.9 MEDIANS AND PERPENDICULAR BISECTORS 



In Figure 5-18, M is the midpoint 
of side W of A ABC Segment AM is 
called a median of A ABC. Since each 
side of a triangle has exactly one mid- 
point, every triangle has exactly three 
medians. Draw a triangle and its me- 
dians. What property do the medians 
appear to have? 




Figure .MS 



Definition 5.5 A median of a triangle is a segment whose 
endpoints arc a vertex of the triangle and the midpoint of the 

side opposite that vertex. 



THEOREM 5.3 The median to the base of 

an isosceles triangle bisects the vertex angle 
and is perpendicular to the base, 

restatement: In A ABC (Figure 5-l c r, 
AB == AC and AM is the median to BC. 

Prove: L AM is the midray of Z BAC. 
2. AM 1 W. 

Proof: Copy and complete the following proof. 




1. 7&mM 

2. ZBs= LC 

3. AM is the median to EC 
4* M is the midpoint of EC, 

6. AABM^AACM 

7. ZBAAf == /.CAM 

8. Af is in the interior of Z BAC, 

9. AA? is the midray of L BAC. 

10. /.BMA= I CM A 

11. IBMA and Z.CMA are a 
linear pair. 

12. Z BAM and Z CAM are right 
angles. 

13. AM _L BC 



1. Hypothesis 

2. \t}{\2}) 

3. Hypothesis 

4.0(1 

8. Theorem 4.11 (4) 

9. Definition of midray (7, 8) 

io. 03 (00) 

11. Definition of linear pair (4) 

12. Theorem 4.12(10, 11) 

13. 0(0) 



5,9 Medians and Perptndicular Bisector* 

In Figure 5-20, A ABC is isosceles with vertex angle at A. Ray AG 

is the midray of the vertex angle. It appears from the figure that AG 
bisects BC and is perpendicular to EC at M. This brings us to our next 
theorem. 



235 




Figure 5-20 

THEOREM 5.4 The midray of the vertex angle of an isosceles 
triangle bisects the base and is perpendicular to it. 

Proof: In proving this theorem it is necessary to show that ray AG 
intersects BS in a point M that is between B and C (as Figure 5-20 sug- 
gests), By the definition of midray, AG is between rays A/5 and AC. By 
the definition of betweenness for rays, B and C are in opposite half- 
planes with edge AG. By the definition of opposite sides of a line, there 

is a point Af of AG between B and C, Since M is between B and C, it 
follows from Theorem 4.11 that M is in the interior of L BAC. Since 

M is in the interior of L BAC and since it is a point of AG or of opp AG, 

but not both, it follows from Theorem 4. 1 that M is a point of AG. The 
rest of the proof of Theorem 5.4 is straightforward and is left as an 
exercise. 

In Theorem 5.4, it was required to prove that a certain ray was per- 
pendicular to a certain segment and that the ray bisected the segment 
The line that contains the ray and that is in the same plane as the seg- 
ment is called the perpendicular bisector of the segment in that plane. 
We state this formally in the following definition. 



Definition 5.6 The perpendicular bisector of a segment in 
a given plane is the line in that plane which is perpcndiculat 
to the segment at its midpoint. 



236 Congruence of Triangles 



Chapter & 



In Figure 5-21, line I is perpendicular to A73 at M t the midpoint of 

A~B. Since I and AB determine exactly one plane (Why?), we say that / 
is the perpendicular bisector of AB in that plane. We sometimes write 
"I — bisA~B" for "/ is the perpendicular bisector of A~B." 



M 



Figure 5-21 

Note that, in space, there is more than one line (How many?) that 
is a perpendicular bisector of a given segment. However, in a given 
plane the perpendicular bisector of a segment is unique, since a seg- 
ment has exactly one midpoint and, by Theorem 4.14, in a given plane 
there is exacdy one line that is perpendicular to a given line at a given 
point on the line. The next two theorems serve to characterize the set 
of points in the perpendicular bisector of a segment, 

THEOR&W 5.5 (The Perpendicular Bisector Theorem) If, in a 

lhvcu plant* a. Pis a nnitil on the perpendlCMbl I fceGtOE of AHk tfagB 
P is equidistant from the endpoints of AR 

restatement: In Figure 5-22, P is a point on line / in a plane a. 

I JL bisAB at M in a, l 

Prove: PA - FB 

Proof: If P = M, then PA = PB 
by the definition of midpoint. If P 
is any point in I different from M, 
then &APM £= ABFM by S.A.S. 
{show this); hence PA == ?B by the 
definition of congruent triangles 
and PA = PB by the definition of 
congruent segments. 

Figure 5-22 




5.9 Medians and Perpendicular Bisectors 237 

THEOREM 5,6 (Converse of the Perpendicular Bisector The- 
orem) If, in a gpven plane a , P is equidistant from the endpoints 
of AB t then P lies on the perpendicular bisector of XB, 
RESTATBMKNT! In plane a, PA s PB, M is the midpoint of AB. 
Prove: P is on I, the perpendicular bisector of AB. 

l 




Proof; If P is on line AB, dien P = M tacause AB has only one mid- 
point, In this case, P is on line I by the definition of the perpendicular 

bisector of a segmen t, I f P is not on line AB, then AAPM =s A BPM by 

S.S.S. (show this). Therefore £AMP is a right angle (Why?) and PM 
± bis AU. Since, in a given plane, a segment has only one perpendicu- 
lar bisector, P is on I, 

EXERCISES 5.9 



L Copy and complete the proof of Theorem 5.3. 

2, Complete the proof of Theorem 5,4 by writing it in two-column form. 

You may assume that it has been proved thai AG intersects J5C in a point 
M which is between B and C, (See Figure 5-20.) 

3. Copy A ABC below. 

(a) Construct the median from A to ISC. 
(b,i On the same figure construct 

the midray of L BAC. 
(c) Does the midray contain the 

median? 

{d) What must be true about 
A ABC if the midray from A 
is to coincide with the median 
from A? 




238 Congruence of Triangles 



Chapter 5 




4. Copy A DEF in the figure. 

(a) Construct the perpendicular 
bisector of EF in plane DEF', 

(b) Does the perpendicular bi- 
sector of EF contain point D? 

(c) What must be true about 
ADEF if _the perpendicular 
bisector of EF is to contain D? 

5. From which theorem may we deduce that the vertex of the angle op- 
posite the base of an isosceles triangle lies on the perpendicular bisector 
of the base? 

0* In the figure. A, B, C, D are distinct coplanar points, KB St /03, and 

(a) Is BD tile perpendicular bisector of AC ? 

(b) Is AC the perpendicular bisecLor of BD? 

(c) Which auxiliary segment is needed to 
prove AABCm AADCf 

(d) How do we know this segment exists? 

(e) Why is IB^ID? 

(f) If your answer to (a) is Yes, prove it is 
correct with a paragraph style proof. 
Do not use congruent triangles, 

(g) If your answer to (b) is Yes, prove it is 
correct with a paragraph style proof. Do 
not use congruent triangles. 

7. In the figure, / is the perpendicular bisector of FQ. If the lengths of .seg- 
ments are as marked, find a, h, c, and d. 








12/ / 

/ V 


\J \ 


K 8 


c \ 


rf%. 


y^ 







5.9 Medians and Perpendicular Bisectors 239 

I In the figure, A ABC is isosceles with AR = BC. If BD is a median, 
prove that AABD s ACBD. 




h In the figure, A PQR is isosceles with PQ b HI, The midrays of L QRP 

and L PQR intersect at T. Prove thai P? 1 QR. Do not use congruent 
triangles in your proof. 




10. If t _L bis A~B at M, and if P is a point on / different from M, prove that 
FKI is a median of A PAR. 

11. If S is the midpoint of (JR andPS 1 pR, prove that APQR is isosceles. 
(It. is not necessary to use congruent triangles in your proof,) 

12. Given the figure as marked with F the midpoint of A~B, prove that 
DFl IE 




13. < itALLtvGt problem. If AB and ($R are coplanar and congruent seg- 
ments, n 1 btsA~Q, m 1. bis BR, and mC n = P, prove that AABP =* 
AQRF. 



240 Congruence of Triangles 



Chapter 5 



In Exercises 14-22, you will need to prove certain triangle* congruent that 
arc not necessarily coplanar. You should look at the figures in perspective 
and be aware that angles or sides that are congruent may not look congruent 
in the figures. In every case, carefully draw a figure on your own paper and 
mark on it the congruent parts before writing a proof. 



14. In the figure, points A, B, and C 
are in plane a. Point P is not in 
plane a. TB _ AB and FB 1 BC. 
If L BAC = £BCA t prove A~P ss 
UP, 




15, In the figure, points D, E, and F 
are in plane /?. Point R is not in 
plane £ If RD jl DE f KD _ DF, 

and DE^W, prove A REF is 
isosceles and that LREF ^ 
£RFE. 




16. In dihedral angle A -RS H. AUis in plane a, BC is in plane /?, ZC ± iT5 r 
HC i RS. D is in /IS. and \A('.B is isosceles with vertex stl C. Prove 
that A BDA is isosceles. 




5.9 Medians and Perpendicular Bisectors 241 
17. Gives TR, W, and KT bisect each other at S. Prove AKDP =± AT/R 

A 





^— ^ 



18. In dihedral angle E-AB-F, E is in plane ft, K is in plane /J, and AB con- 
tains C and D, If triangles DEC and DFX.' are isosceles with vertex an- 
gles at £ and F, respectively, prove that L EOF == Z ECF, (Plan: Prove 
AEDF s AECF by S.S.S.) {See figure above at right) 

19. In Exorcise 18, is DC l>etween rays DF and DE? Can we prove 
Z.EDF& Z ECF by the Angle Measure Addition Postulate? 



20. In the figure, PA = PB = 
PC and lAPB=zlBPC=z 

I A PC. Prove that AABCte 
equilateral. 



21. In Kxercise 20, if we accept the hypothesis (but disregard the figure}, is 
it possible that P, A t B> C are coplanar? Try to draw a figure for the 
"plane case," 

22. caiAjxENCE problem. In the fig]:. /' ;iru.l pare on opposite sides of 
plane a which contains points D, E, and EUPQ ±DK at E,TQ ± FE 
at E, and M == @E, prove &FDF ^ A QDF. 





242 Congruence of Trfangles Chapter 5 

CHAPTER SUMMARY 

The Definitions 

If t for A ABC and ADEF. 

L A *. — » ZD, KB < — » DE, 

IB < — * IE, W < — > EF, 

I C * — ► IF, AC < — > JJF y 

we indicate this correspondence by writing ABC <—— * DEF where the or- 
der in which the vertices arc named preserves the six correspondences. We 
speak of the pairs In the six correspondences named as CORRESPONDING 
PARTS of the two triangles. Two triangles (not necessarily distinct) arc 
CONGRUENT if and only if there exists a one-to-one correspondence be- 
tween their vertices in which the corresponding parts arc congruent Such 
a one-to-one correspondence between the vertices of two congruent trian- 
gles is called a CONGRUENCE. 

An ANGLE of a triangle is said to be INCLUDED by two sides of that 
triangle if the angle contains those sides. A SIDE of a triangle is said to be 
I N( .1 .UDED by two angles of that triangle if the cudpoints of the side are 
the vertices of those angles. 

The combination TWO SIDES AND THE INCLUDED ANGLE is 
abbreviated by the symbol S. A«S, If there exists a correspondence between 
two triangles such that S.A.S. of one triangle are congruent to S.A.S, of the 
second triangle, then we call this an S.A.S. CONGRUENCE, 

The combination TWO ANGLES AND THE INCLUDED SIDE is 
abbreviated by the symbol A.S.A. If there exists a correspondence between 
two triangles such that ASA. of one triangle are congruent to ASA, of the 
Second triangle, then we call this an ASA. CONGRUENCE. 

The combination THREE SIDES is abbreviated by the symbol S.S.S. 
If there exists a correspondence between two triangles such that S.S.S. of 
one triangle are congruent to S.S.S. of the second triangle, then we call this 
an S.S.S. CONGRUENCE. 

If a triangle has (at least) two congruent sides, it is called an ISOS- 
CELES triangle. The VERTEX ANCLE of an isosceles triangle is the angle 
included by the congruent sides. The BASE of an isosceles triangle is the 
side that is opposite the vertex angle. The BASE ANGLES of an isosceles 
triangle are the angles whose vertices are the endpolnts of the base. An 
EQUILATERAL triangle is a triangle that has three congruent sides. An 
EQUIANGULAR triangle is a triangle that has three congruent angles. A 
MEDIAN of a triangle is a segment that has for its endpoints a vertex of the 
triangle and the midpoint of the side opposite that vertex, 

In a given plane, the PERPENDICULAR BISECTOR OF A SEG- 
MENT is the line that is perpendicular to the segment at its midpoint. 

A statement of the form "If fh then </" is called a CONDITIONAL. 
The statement p is called the HYPOTHESIS of the conditional and the 



Chapter Summary 243 

statement q is called the CONCLUSION. If a conditional and its hypothesis 
are known to be true, then it follows logically that its conclusion is also true. 
The statement **ff q, then p" is called the CONVERSE of the statement 
"If p, then q" and the statement "If p t then q" is called the converse of the 
statement "If q y then p" A definition in "if-then" form is to be understood 
as a conjunction of two statements "If p, then q" and "If q t then p" Some- 
times this is abbreviated to "p if and only iiq," 

The Postulates 

There were three postulates in this chapter having to do with congru- 
ent triangles. We list them In their abbreviated form only. Be sure that you 
know the complete statement of each postulate. 
POSTULATE 23. The S,A<S. Postulate. 
POSTULATE 24. The \SJi. Postulate, 
POSTULATE 25. Tlie S.S.S. Postulate. 



The Theorems 

There were six tlieorems and two corollaries in this chapter. We list 
them in the order in which they appeared. Re sure that you know what each 
theorem says and that you understand its proof. 

THEOREM 5 J (77ie hosceles Triangle Theorem) The base angles 

of an isosceles triangle are congruent. 

COR OLLAR Y 5.1.1 If a triangle is equilateral, then it is equiangular. 

THEOREM 5.2 (Converse of the I*o$eek* Triangle Theorem). If a 
triangle has two congruent angles, then the sides opposite these angles 
arc congruent and the triangle is isosceles, 

COROLLARY 5,2. 1 I! a Lrifuigle is equiangular, then it is equilateral 

THEOREM 5.3 The median to the base of an isosceles triangle bi- 
sects the vertex angle and is perpendicular to the base* 

THEOREM 5.4 The midray of the vertex angle of an isosceles tri- 
angle bisects the base and is perpendicular to it. 

THEOREM 5.5 (The Perpendicular Bizector Theorem) If, in a 
given plane a, P is a point on the perpendicular bisector of KB, then P 
is equidistant from the endpofnts of KB. 

THEOREM 5.6 (Converse of the Perpendicular Bisector Theorem) 

If, in a given plane «. P is equidistant from the endpoints of KB, then 
P H*s on die perpendicular bisector of KB. 



244 Congruence of Triangles Chapter 5 

REVIEW EXERCISES 

1. Write the fallowing statement as a conjunction of two statements in the 
"if-then" form: A triangle is equilateral if and only if it is equiangular, 

2. In your answer to Exercise 1, what is the second statement called with 
respect to the first? Are both statements true? 

3. Write the converse of the statement: If two triangles are congruent, 
then their corresponding parts are congruent 

4. In Exercise 3, are both the statement and its converse true? 

5. Write the convene of the statement: Vertical angles are congruent. 

6. In Exercise 5. is the given statement true? Is its converse true? 

■ In Exercises 7 IS, decide if the sentence is true or false. 

7. In the statement, "If p, then q" p is the conclusion and q is the 
hypothesis. 

8. In the statement, "p only if q," p is the conclusion and q is the 
hypothesis, 

9. In the statement, "p if tf," p is the conclusion and q is the hypothesis. 

10. If we know the statement "If jt>, then q" is true and if we know that p 
is true, then we can conclude that q is true. 

11. TJie statement "If p, then qr" may be true even though the statements 
p and q are both false, 

12. If A is on the perpendicular bisector of CD, then AC = AD, 

13. If, in a given plane, PE = PF, then P is on the perpendicular bisector of 

EF. 

14. In a given plane, more than one line can be drawn which is both per- 
pendicular to a segment and bisects the segment. 

15. In a given plane, more than one line can be drawn which bisects a 
segment. 

16. If R, S t and T are not colLuiear and R is on the perpendicular bisector 
of ST, then A RST is isosceles, 

17. Congruence of triangles has the transitive property. 

18. In our formal development of geometry, we postulated that the sides 
Opposite a pair of congruent angles in a triangle are congruent. 

19. Copy and complete: If, in a given plane, R is (?] from the endpoints of 
ST. then R lies on the (7] [3 of ST. 

20. Copy and complete: If JAB is a segment and if R and S are disti n ct points 
such that R, S, A, H are coplanar, RA = RH, and M = SB, then jTj is 
the |7]|l] of A3- 

21. If, la A ABC, ED bi sects Z ABC and intersects AC at D, ~KB m EC, 
prove that D in the midpoint of AC. 



22, If, in the figure, A~B - BC, does 
it follow that lBADs*£BCD 
because base angles of an isosceles 
triangle are congruent? Why? 

23, In the figure for Exercise 22, if 
■\r. w. I BAD a . BCD. and 

D is in the interior of AABC t 
write a plan for proving A ABD ^ 
ACfJD. 

24, Given: AC^BC 

lACD^ Z.BCE 
A-D-Ejy-E-B 
Prove; CD rr CK 



Revktw Exsrcises 245 
B 



25. Given: 



I'rnt r; 



A-D-C, 
B-E-Q B-P-D 
A-P-E, 11? EM RE 
LCDP& I CEP 
AAFB is 
isosceles. 




26. Given: A3 = CD t A-M-Q 
B-XS-D.aBdADssEC 
Prove: (a) A ABD ss ACDB 

(b) ^lsZ2 

(c) A A CD ss A GAB 
fd) Z3aZ4 

(e) ACMD as AAJlfB 

(f) MJs the midpoint of 
BDandAT, 

27. In the plane figure, \f is the 
midpoint of segments A~C and 
EP, BE sm Prove 

(a) AAMFss&CME. 

(b) AABC== ACDA. 

28. Given: A-D-E, D-E-B 

AD"? BE 

CD s ce 
Prove: A ABC is isosceles. 
Prove, in order, 

Zl = /2 
Z3= i. J 
AACDj^ ABCtf 
AC m BC 

A ABC is isosceles. 






Chapter 




Carl Stru wc/Monkmetjtfr 



Inequalities 
in Triangles 



6.1 INTRODUCTION 

In Chapters 3 T 4, and 5 we developed the concept of congruence 
for segments, angles, and triangles. When we say that two segments 
are congruent, we mean that two numbers associated with these seg- 
ments, their lengths, are the same. The concept of congruence for an- 
gles and for triangles is based also on the concept of equal measures, 
and hence on the concept of equality for numbers. 

In this chapter we are concerned with segments and angles, in par- 
ticular with segments that are not congruent to each other and with 
angles that are not congruent to each other. In other words, wc are 
concerned with segments of unequal measures and with angles of un- 
equal measures. In comparing two segments {or angles) that are not 
congruent to each other, it is useful to know which one has the larger 
measure. To express such a comparison it is convenient to use some fa- 
miliar words and symbols defined formally as follows. 

Definition 6.1 XB > CD if and only if AB > CD< AB < 
CD if arid only if Afl < CD. 

Definition 6.2 I ABC > LDEFii and only if m I ABC > 
mLDEF, IABC< IDEF if and onlv 'if mlABC< 
mLDEE. 



248 Inequalities in Triangles Chapter 6 

When used to compare segments, the symbol "]> M may be read as 
"greater than," or "longer than," or "larger than." Similarly, **<" may 
he read as 'less than," or "shorter than," or "smaller than." 

When used to compare angles, the symbol "]>" may be read as 
"greater than" or "larger than." Similarly, "<*' may be read as "less 
than" or "smaller than," 

Note that Definitions 6, 1 and 6.2 express a comparison of segments, 
or of angles, in terms of an inequality involving numbers. It should be 
clear that an inequality involving segments or angles is really a state- 
ment of "not-congruence" together with a statement of which seg- 
ment, or angle, has the greater, or lesser, measure. Because the proper- 
ties of inequalities for numbers are essential for the development of in- 
equalities for segments and angles, we review them in Section 6,2. 

Tliis chapter includes several important theorems, some of which 
involve comparisons of parts of one Lriangle. Others involve compari- 
sons of parts of one triangle with parts of another triangle. To save time 
we shall, in some cases, give "abbreviated proofs" of these theorems, 
that is, just the key steps in the proofs. You should be able to supply a 
complete proof if asked to do so. 

6.2 INEQUALITIES FOR NUMBERS 

We begin by defining* formally, less than and greater than for 
numbers. 



Definition 6.3 If a and b are numbers, then a K h if and 
only if there is a positive number p such that b — a + p. 
Also a > b if and only if there is a positive number p such 
that a = b + p. 

Definition 6.4 If a and b are numbers, then a < b if and 
only if a < b or a = b. 

Example 1 The statement 4 < 5 is read "4 is less than or equal to 5" 
and, by Definition 6.4, it means 4 < 5 or 4 = 5. We know that a dis- 
junction of two statements is true if either of the two statements is true. 
Therefore 4 < 5 is a true statement because 4 < 5 is true. Similarly, 
5 < 5 (which means 5<5or5 = 5)isa true statement because 5 = 5 
is true. 

Most of the numbers in our geometry are positive numbers and 
represent lengths of segments, measures of angles, measures of plane 



6.2 InequalJttot for Numbers 249 

regions (areas), or measures of regions in space (volumes). From Defi- 
nition 6.3 we can conclude that if x, y, and * are positive numbers and 
if x = if + z, then both tj and z are less than x. We get this by first con- 
sidering z as the positive numt>er p in the definition and then consider- 
ing y as the positive number p. This gives us Lhe two statements 

x = y + p and x — z 4- p« 

Thus, by Definition 6.3, y < x and z < x. That is, both numbers in a 
sum of two positive numbers are less than the sum. For example, 

15 = 9 + a 

Hence 9 < 15 and 6 < 15. From the second part of Definition 6.3, we 
can conclude that the sum of two positive numbers is greater than 
either of them. For example, 

32 = 21 -J- 11, 

Heneo32>2J and 32 > 11. 

The following theorem is easy to prove using Definition 6.3. 

THEOREM 6.1 If x and y are numbers, then x < y if and only if 

There are two parts to Theorem 6.1, 

1 . If x and y are numbers, and if x < y, then y > x. 

2. If x and y are numbers, and if y > x, then x < y. 

Proof of part b 

Statement Reason 

1. x and y are numbers and 1. Hypothesis 
x<y. 

2. y = x + p, where p is a posi- 2, Definition 6.3 
tive number. 

3. y > x 3. Statement 2 and Definition 

6,3 

JVoo/ of part 2: Assigned as an exercise. 

We now state five properties of order (inequalities) that are helpful 
in proving theorems about geometric inequalities. You may consider 
these properties as postulates for the real number system, although we 

could prove Properties 0-3, 0-4, and 0-5 by using Definition 6,3 and 
Properties Q-J and 0-2. 



250 ln«qu*fties in Trianglss Chapter 6 

0-1 The Positive Closure Property. If x and y are numbers, and 
x > and y > S then x + t/ > and xy > 0. 

0-2 Trichotomy Properly. If x and t/ are numbers, then exactly 
one of the following is true: x < y> x = I/* x > j/, 

0-3 Transitive Property. Tf a;, y, and z arc numliers, and if x <C J/ 
and |( < # then x < z. Also, if * > y and y > £, then % > % 

0-4 Addition Property. If a, h, x, and y are numbers, and if 
x <C ^ and a < fe, then i + fl < y t &• Also, if x ]> t/ and a > i>, 
then x + a > y -+- fc. 

0-5 Multiplication Property. If x, y, and z arc numbers and if 
x < y and s > 0, then xz <C ys. Also, if x >■ y .and z > 0, then 

We now prove several theorems which are useful in the sections 
that follow. 

THEOREM 6.2 AS > CD if and only if CD < "KB, 

There are two parts to Theorem 6,2. 

1. If AB > CD, then £75 < "KB. 

2. If CD < AB r then IB > CD. 

Proof of part 1: 

Statement Reason 

1. £B>CD L Hypothesis 

2. AB > CD 2. Definition 6.1 
a CD<AB 3. Theorem 6.1 
4. CD < ZB 4. Definition 6. 1 

Proof of part 2: Assigned as an exercise, 

THEOREM 6.3 IABC> ADEF if and only If 
IDEF<: I ABC. 
Proof; Assigned as an exercise. 

THEOREM 6.4 Let three distinct collinear points A, B, C be 
given. Then A-C-B if and only if AB > AC and AB > BC 



6.2 Inequalities for Kumbers 251 

Tnere are two parts to Theorem 6.4. (Draw a figure showing the re- 
lationship between points A, B, and C.) 

J, Given three distinct collinear points A, B ? C, if point C is be- 
tween points A and B, then AB > AC and AB > BC. 

2. Given three distinct collinear points A, B, C, if AB > AC and 
AB > BC^ then C is between A and B. 

Proof of part ! . Since C is between A and B, AB = AC -f BC, Why? 
But AC and BC are positive, Hence by Definition 6.3 AB > AC and 
AB > BC. % 

Proof of part 2; Points A. B, C are coUinear (given), so exactly one of 
them is between the other two. Why? That is, we must have exactly 
one of the following three belweenncss relations: B-A-C t A-B-C, or 
A-C-B. We shall show that the first two of these are impossible; hence 
the only remaining possibility is A-C-B. Suppose that A is l>etween B 
and C, Then, by the first part of Theorem 6,4, we must have BCy AB. 
But this contradicts the hypothesis that AB > BC; hence A is not be- 
tween B and C. Now suppose B is between A and C. Again, by part 1 
of Theorem 9.4, we must have AC > AB and this contradicts the hy- 
pothesis that AB > AC; hence B is not between A and C. The only 
remaining possibility is that C is between A and B. This completes the 
proof. 

THEOREM 6.5 If point D is in the interior of I ABC, then 
mlABC>mlABD and m£ABC> m£DBC. 



Proof: Since D is in the interior of /.ABQ ray BD is between rays 
BA andBC. By the Angle Measure Addition Postulate, 

mLABC = mlABD + m£DBC> 

Since mLABD and mLDBC are positive, it follows from Definition 
6,3 that 

m Z ABC > m Z ABO and m Z ABC > m Z DBC. 

EXERCISES 6.2 

In Exercises 1-10, identify the order property that is illustrated. 

L If AB < 6. then AB # a 

2. If a - fe < 15 and ft < 3, then a < 18, 

3. If x < 7 and 7 < y, then x < y. 



252 Inequalities in Triangles 



Chapter 6 



4. If a < 5, ihen 4o < 20. 

5- If AB < RS and BC < ST. then AS + BC < HS + ST, 

6. If Jm Z ABC > ^m Z RST, then m Z ABC> m £ R ST. 

7. If Ali > CD and CD a EF, then AB > EF. 

8. If x + 3 < 8, then a: < 5. 

9. If x - y > 12 and y = 7, then x > 19. 

10. If 3 > and 2 > 0, then 2 + 3 > 0. 

11, in the figure, AD > BE and DC > EC. 
Prove AC >' BC. 



12. Given the figure for Exercise 11, if 
m £ 1< m £2 and m Z3 < to Z4 
prove that m Z ABC > m £ BAC. 



13, Ii m/.A = 90 + k, where k > 0, and ZB is u supplement of ZA, 

prove that Z 2? is an acute angle. 

14. In the figure, £D =z £ DEC. 15. In the figure, CD JL bis AB and 
Prove that m Z ABC > ffl Z D. C-P-B. Prove that AC > CE. 






16. Prove part 2 of Theorem 6.1, 

17. Prove part 2 of Theorem 6.2. 

18. Prove Theorem 6.3. 

19. Explain why Theorem 6.4 has the following consequence. If A, B, C 

are three distinct collinear points, then C is between A and B if and only 
if AC <A£ and BC< AB. 

20. Explain why Theorem 6.5 has the following consequence. If D is a point 
in the interior of /ABC, then 

m I ABD < to £ ABC and m £DBC < m £ABC. 

P 

21. In the figure, M is the midpoint 
of AT and BC. Prove tliul 

m£BCD> ml B. 




6.2 Inequality for Number* 253 

22. For each figure use your protractor to measure Z BCD y LA, and L B, 
Record your results in a table. How does m/ BCD compare with mLA 
in every cu.se? m Z BCD with m L B in every ease? 



m BCD 



m A 



mLB 




254 Inequalities in Triangfes 



Chapter 6 



23. challenge problem. Gi vet i the coplanar potato A,B,C, D, J#, P such 
that A, B, C are noncollincar, A~C~D, B-M-C, and A-M-P, prove that 
P is in the interior of L BCD. (Hint: You must show that P is on the 

tf-side of S/5 and that P is on the D-side of BC.) 

24, challenge pbohlem. Prove the Transitive Property of Order (Prop- 
erty 0-3). {Hint: You will nocd to use Definitions 8.3 and 6.4 and the 
Addition Property of Kquality in your proof.) 

6.3 THE EXTERIOR ANGLE THEOREM 

In both figures of Figure 6-1, ZABC, ZBGA, and LCAB are 
called interior angles of A ABC, We call /.BCD, which is adjacent 
to LACB and forms a linear pair with it, an exterior angle of A ABC. 





Both Z A and Z B are called nonadjacent interior angles of the exterior 
angle Z BCD. The term "interior angle" is convenient when you want 
to emphasize the distinction between an exterior angle of a triangle and 
an angle of a triangle. \ T ote diat the adjectives "adjacent" and "non- 
adjaeerit" apply to an interior angle and descrrt>e its relation to a par- 
ticular exterior angle. We formalin these ideas in die following 
definition. 



Definition 6.5 Each angle of a triangle is called an interior 
angle of the triangle. An angle which forms a linear pair with 
an interior angle of a triangle is ealled an exterior angle of the 
triangle. Each exterior angle is said to be adjacent to the in- 
terior angle with which it forms a linear pair and nonadjacent 
to the other two interior angles of the triangle. 



Every triangle has six exterior angles, two at each vertex, as shown 
in Figure 6-2, The two exterior angles at each vertex are vertical angles 
and hence are congruent. 



6,3 The Exterior Angle Theorem 255 




Figure fi-2 

Note that L DCE in Figure 6-2 Is not an exterior angle. Why? In 
Figure 6-2 T L KAC and I HAB are the two exterior angles at vertex A 
of &ABQ and LABC and LBCA are the two nonadjacent interior 
angles of each of them. Name the two exterior angles at vertex B of 
A ABC in Figure 6-2 and their nonadjacent interior angles. 

If you worked Exercise 22 of Exercises 6,2, your results should sug- 
gest the following theorem. 

THEOREM 6.6 {The Exterior Angle Tfteorem) Each exterior 
angle of a triangle is greater than either of its nonadjacent interior 
angles. 

Proof: Let the vertices of the triangle be A . B, and C. T iet D he a point 
on AC such that C is be- n p 

tween A and D. (See Figure /S. ""/ 

6-3.; We must prove that / X. 

LBCO> LBAC 
and that 

LRCD> LB. 
We first prove that j**- 

IBCDy LB. Figured 

Let M be the midpoint of BC and let P be the point on AM such that 
A-M-F and AM = MR Then AAMB == A PMC by S.A.S. (show this) 
and mLBCP = mLB. Why? Since P is in the interior of /BCD, 
m L BCD > m L BCP by Theorem 6.5. We have shown that 

mLBCD > mLBCP 
and that 

mLBCP- mLB. 

It follows from the Transitive Property (0-3) that mL BCD > mLB, 
and from Definition 6.2 we have LBCD > / B. 




256 Inequalities In Triangles Chapters 

To prove that Z BCD > Z BAG, we use the midpoint of AC and 
show that the other exterior angle at C {LACE in Figure 6-4) is greater 
than / BAG in the same way as in the part above. Since the two ex- 
terior angles at C are congruent (Why?), it follows that 

IBCD> I BAG 




Figure <M 

The proof that we have given for one exterior angle of the triangle 
can be easily modified to show that die theorem holds for any of the six 
exterior angles of the triangle. However, it is not necessary to go to all 
this trouble since our choice of the exterior angle at C was stricdy an 
arbitrary choice. The fact that we were able to prove the theorem by 
choosing arbitrarily any one of the six exterior angles of A A /?C insures 
us that there is no need to prove the theorem for every exterior angle. 

In the proof of Theorem 6,6\ we stated that point P (in Figure 6^3) 
is in the interior of Z BCD. You are asked to prove this in the Exercises 
at the end of this section. 

COROLLARY 6.6, 1 If one of the angles of a triangle is a right 
angle, then the other two angles of the triangle are acute angles. 

Frvof: Assigned as an exercise. 

COROLLARY 6.6,2 If one of the angles of a triangle is an obtuse 
angle, then the other two angles of the triangle arc acute angles. 

Proof: Assigned as an exercise. 

It follows from the Exterior Angle Theorem (more directly from 
its corollaries, Corollary 6.0.1 and Corollary 6.6.2) that no triangle has 
more than one right angle or more than one obtuse angle. An important 
kind of a triangle is one that has one right angle. There are special 
names for triangles with a right angle and for triangles with an obtuse 
angle. 



6.3 The Exterior Angle Theorem 257 



Definition 6.6 A right triangle is a triangle with one right 
angle. The hypotenuse of a right triangle is the side opposite 
the right angle. The other two sides of a right triangle are 
called legs. 

Definition 6.7 An obtuse triangle is a triangle with one ob- 
tuse angle. 

Definition 6.8 An acute triangle is a triangle with three 
acute angles. 



Theorem 4.14 asserts that for each point on a line in a plane, there 
is one and only one line which (1) lies in the given plane. (2) contains 
the given point, and (3) is perpendicular to the given line. We can now 
use the Exterior Angle Theorem to prove a companion theorem. 

THEOREM 6, 7 Given a line and a point not on the line, there is 
one and only one line which contains the given point and which is 
perpendicular to the given line. 

Proof: Let I be the given line and P the given point not on L In part 1 
of the proof we show that there is a line containing P and perpendic- 
ular to L In part 2 we show there cannot be two such lines. 

1. Existence. (See Figure 6-5,) Let A and B be any two points of 

line I. Then PA is either perpendicular to / or not perpendicular to /. 

If FA ± h then the existence part of our proof is complete. If PA is not 

perpendicular to /, then there is a ray AC* with C on the opposite side 
of / from P such that I PAB == 
ABAC. Why? There is a point D 

on AC such that AD at Aft Why? 
P and C are on opposite sides of /, 
and C and D arc on the same side 
of l\ hence P and D are on opposite 
sides of /. Therefore PD intersects 
/ at some point F. APAF^ &DAF 
by SJLS. (shew tills} and so 
L PEA as LDFA. Therefore, by 
Theorem 4.12, LPFA is a right 

angle and PF ± I 

Figure &S 




258 Inequalities in Triangles 



Chapter 6 



2. Uniqueness. (See Figure 6*6.) Let C be any point of / other than 
F and let H be any point of / such that G-P-H. Then Z PCFis an angle 

of APGF and is a nonadjacent * 



interior angle of the exterior 
angle Z PFIL Since m L PFH = 
90, it follows from the Exterior 
Angle Theorem that mlPGF 

< 90 and therefore PG is not 
perpendicular to L It follows 

that PP is the only hue through 
P and perpendicular to L 




Figure fi-fi 



EXERCISES 6.3 



1. Refer to Figure 6-4 and prove that Z BCD > Z MCj thus completing 
the proof of Theorem 6,6. 

2. Prove Corollary 0.0. L 

3. Prove Corollary 6.0.2. 

4. Given A ABC with B-C-D and mLACD = 70, what must be true 
about m LABC? About r« Z BAG? About mAACB? 

5. In the figure, name the two 
nonadjacent interior angles of 
ZBAP "Which exterior angle 
has A CAB and I ABC as its 
nonadjacent interior angles? 
Name all the exterior angles 
shown in the figure and their 
corresponding nonadjacent inte- 
rior angles. 

Using the figure, copy Exercises 6- 1 1 and replace the question marks by 
<, = t or > to make a true statement, a, b t c t x t y._ z denote angle measures. 





6.3 The Exterior Angle Theorem 259 

6. If b = 40, then x [?] 40 and y [?] 140, 

7. If c = 90, then z \t] 90, a |Tj 90, b Q] 90, x (?)90, and y |7]90. 

8. If a = 40 and b = 60, then 2 f7] GO. 

9. If (/ = 140, then a [T] 140 and c [7| 140. 
10. If x = 130 t then h [?] 130. 

1L If a = 55 and c = 80, then i/ [?] 80. 

Refer to the figure for Exercises 12-18. In each exercise, arrange the nuni- 
bers in order, starting with the smallest. In these exercises, <i, h t c> d, e,f t g 
denote angles. 




12. m La, m Z c 

13. mLh t mLd 

14. mle, mZc 

15. mZb» BlZ/ 



16. m £c ml_a t mLe 

17. mZc, m Z e, rn Z g 

18. mZf ( mZa,»nZg 1 »nZ« 



19. In the Bgure, prove that mlACB > mLB. 

A 




20. List the angles marked in the figure In Increasing order of size. 




260 Inequalities in Triangles 



Chapter6 



21 ♦ In the proof of Theorem 6,B it was stated that the point P (Figure 
was in the interior of L BCD, To prove this* wc must show that P is on 
tile D-sidc of S3 and on the B-side of c3. 




A C 

Copy and supply the missing reasons iu the following proof. 

Statement 



A-C-D, A-M-P, B-M-C 

M and F are on the same 

side of S3. 

M and B are on the same 

side of 53. 

P is on the B-side of CD. 
A and D are on opposite 

sides of BC. 

A and P are on opposite 

sides of BC, 

P is on the D-sidt of BC, 
P is in the interior of 

/ BCD. 



L Given 

2, Theorem 2.2 

3. m 

4. Statements 2 and 3 

5. \?} 

7, Statements 5 and 

8. EKHHIl) 



22. Is it possible for a triangle to have two right angles? Justify your answer. 

23. Suppose that I is a line in plane a and that F is a point not in a. Docs 
Theorem 6.7 still apply? Draw a figure and explain your answer. 

24. Given: AAMC f A-F-S> 25. Given: AADB.A-C-D, 

F-S-C AD = "KB 

Pmw,: Z1>Z2 Prove: LACW> LDBA 





6.3 The Exterior Angle Theorem 261 



h Cwtmt B is the midpoint of 
333, 8 is the midpoint of FC, 

B-c-n 

Prom: IDCE> LA 




27. Gtem: ARTS, R-W-l\ SW 
is the midray of L RST 
Prove; Z3>Z1 



28, Given: Quadrilateral A BCD, 

B-K-C, 3D ss DE t L 1 ~ 

Z2 

Prvce: Z3>Z1 




29. Gkert: F is the midpoint of 
AD and BE 
Prove: Z2<Z1 





30, Measure the sides in centimeters and the dingles in degrees of the scalene 
triangle in the figure. Record the measurements to the nearest tenth of 
a centimeter and the nearest degree in a table as shown. Yon will need 
to refer to the results of this exercise in Section 6.4. 



Angles 



c = f?l 



m/.A=\J\ 
ifiZB = f7| 




262 Inequalities m Triangles Chapter 6 

31. Draw three scalene triangles of different sizes and shapes and label the 
vertices and sides as in Exercise 30. Also make and record measure- 
ments for each triangle. You will need to refer to the results of this ex- 
ercise in Section 6,4. 

32. a r allence problem , If Pis any point in the interior of A A 8C, prove 
that m£APB> mlC. 

33. challenge pkohlem. Prove that the sum of the measures of any two 
angles of a triangle is less than ISO, 



6.4 INEQUALITIES INVOLVING TRIANGLES 

If all three sides of a triangle are congruent, then die three angles 
of the triangle are congruent, and if two sides of a triangle arc congru- 
ent* then the angles opposite these sides are congruent. In this section 
we investigate how the angles of a triangle are related to each other 
when they are opposite Sides of unequal length. 

Refer to the table you prepared in Exercise 30 of Exercises 6.3 and 
answer the following questions. 

Which side is the longest? 

Which angle is the largest" 

How are the longest side and the largest angle of 

AABC situated with respect to each other? 
Which side is the shortest? 
Which angle is the smallest? 
How are the shortest side and the smallest angle of 

AABC situated with respect to each other? 

Observe the order relation among the measures of the angles of 
AABC and complete the following statement; 

mL\2] > mZQ] > mZ|T]. 

Observe the order relation among the measures of the sides of 
AABC and complete the following statement: 

EJ>E>[& 

How do the order relations among the measures of the angles of 
the triangle compare with the order relations among the measures of 
tile corresponding opposite sides of the triangle? 

Refer to the tables you prepared for the three triangles in Exercise 
31 of Exercises 6,3 and answer the same questions as the preceding 
ones for each of these triangles. Make a general statement about the 
relative position of Lhe longest side and the largest angle of a triangle; 
the shortest side and the smallest angle. 



6,4 Inequalities Involving Triangle* 263 

The comparisons suggested by the preceding experiment are for- 
mulated in the following two theorems, 

THEOREM 6*8 {Angle-Comparison Theorem) If two sides of u 
triangle are not congruent, then the angles opposite them are not 
congruent and the greater angle lies opposite the greater side. 

Proof: 

Given; AABC, 

with AB > ~KC 
To From: I ACS > I ABC (See Figure 6-7.) 




Statement 

1 „ AB > £C 

2. AByAC 

3. There is a point D on 
such that KD zzlC. 

4. AD = AC 

5. AB>AD 



6. D is between A and B. 

7. D is in the interior of £.ACB. 

8. mlACB>mAACD 

9. lACDss I ADC 

10. mlACD = mlADC 

11. nZACB > mZ40C 

12. ZADC> ZAJ3C 

13. m£ADC>mLABC 

14. mZACB>mZABC 

15. lACBy IABC 



Reason 



1. Given 

2. Definition 6. 1 



3. Segment Construction The- 
orem 

4. Definition of congruence for 
segments (3) 

5. Substitution Property of 
Equality (2, 4) 

6. Theorem 6,4 

7. Theorem 4,11 

8. Theorem 6.5 

9. Isosceles Triangle Theorem 
(3) 

10. Definition of congruence for 
angles (9) 

11. Transitive Property (8, 10) 

12. Exterior Angle Theorem 

13. Definition 6.2 (12) 

14. Transitive Property (U, 13) 

15. Definition 6.2 (14) 



264 Inequalities in Triages Chapter 6 

We now state and prove the converse of Theorem 6.8. 

THEOREM 6.9 (Side-Comparison TJieorem) If two angles of a 
triangle are not congruent, then the sides opposite them are not 
congruent and the greater side lies opposite the greater angle, 

Proofs Let A ABC be any triangle with two angles that are not con- 
gruent Suppose it has been named so that these noncongruent angles 
are / B and Z C and such that LC> L B, In terms of this situation 
the hypothesis and the conclusion to be proved are as follows. 

Hypothesis: LC> LB 

Conclusion: AB > AC 

(See Figure 6-S.) 




Since AB and AC are numbers, one of the following must hold: 

(1) AB<AC 

(2) AB = AC 

(3) AB > AC 

Which property of numbers are we using here? 

The method of proof is to show that (1) and (2) are impossible, so 
(3) must hold, thus proving the theorem. 

(1) If AB < AC, then, by Theorem 6.8, LC < LB, This contra- 
dicts the hypothesis; hence AB < AC is impossible. 

(2) If AB = AC then A ABC is isosceles and LC ss LB. Again 
this contradicts the hypothesis; thus we see that AB = AC is 
impossible. 

It follows then that AB > AC so that AB > KC and the desired 
conclusion has been proved. 

The following two corollaries follow immediately from Theorem 

6.9 and Corollary" 6.6.1. 

COR OLLAMY 6.9. 1 The hypotenuse of a right triangle is the 
longest side of the triangle. 

Proof: Assigned as an exercise. 



6,4 Inequalities Involving Triangles 265 



COROLLARY 6.9,2 The shortest segment from a point to a line 
not containing the point is the segment perpendicular to the line. 

restatkmeoti Given a Hue I 
and a point P that is not 

on I, if PA J. I at A and 
B is any other point of ?, 
then PA < FB. (See Fig- 
ure 6-9.) 



►J 



Proof: Assigned as an exercise. 



a B 

Figure 6-9 

When we speak of the distance between a point P and ft line l> we 
naturally mean the shortest distance from J' to I. It follows from Corol- 
lary 6,9.2 that there is such a shortest distance, and so we make the 
following definition. 



Definition 6,9 The distance between a point and a line not 
containing the point is the length of the perpendicular seg- 
ment joining the point to the line. The distance between a 
line and a point on the line is defined to be zero. 



It is customary to associate three "distances between ft point and a 
line" with every triangle. With A ABC there is associated the distance 

between A and EC, the distance between B and CA, and the distance 



between C and AS. Any side (or its length) of a triangle may be thought 
of as the base. Associated with each base is the segment (or its length) 
joining the opposite vertex to a point of the line containing the base. 
The following definition has two parts. In (I), we define base and 
altitude, thought of as segments. In (2), we define base and altitude, 
thought of as numbers (lengths of segments or distances). 



Definition 6,10 

L Any side of a triangle is a base of that triangle. Given a 
base of a triangle, the segment joining the Opposite vertex to 
a point of the line containing its base, and perpendicular to 
the line containing the base, is the altitude corresponding to 
tluit base. 

%. The length of any side of a triangle is a base of that tri- 
angle. The distance between the opposite vertex and the line 
containing that side is the corresponding altitude- 



266 Inequalities in Triangle* 



Chapter 6 



Figure 6-10 shows two triangles, A ABC and AA'B'C. In each tri- 
angle the segments from the vertices to the lines containing the oppo- 
site sides have been drawn. These segments are the altitudes of the 
triangles. Thus XD is an altitude of A ABC. It is the altitude from A to 
sideTJC. The point D is the foot of the altitude from A to EC. Note for 
an acute triangle, such as AABC in the figure, that the foot of each 
altitude is an interior point of a side. Note for an obtuse triangle, such 
as A A'B'C in the figure, that the feet of two altitudes ar« • not points of 
the triangle. Even though point E\ for example, is not a point of side 
A'C. it is frequendy called the foot of the altitude from B' to side A^U. 





Figure 640 

Draw a right triangle, A ABC, with the right angle at C. Draw the 
altitude from C to the hypotenuse. Is the foot of this altitude a point 
of the triangle? Name the other two altitudes of A ABC. Are the feet 
of these altitudes points of the triangle? 

In Chapter 3 we postulated that if A, B, C are noncolliuear points, 
then for distances in any system, AB -\- BC > AC. The postulate was 
called the Triangle Inequality Postulate. We now have the necessary 
geometric properties to prove this statement as a theorem. 

THEOREM 6.10 (Triangle Inequality Theorem) The sum of the 
lengths of any two sides of a triangle is greater than the length of 
the third side. 



Proof: Given any triangle, it follows from the Trichotomy Property 
for numbers that there is one side of the triangle which is at least as long 
as each of the other two sides. Suppose, in A ABC, that BC j> AB 
that BC > AC, (See Figure 6-11.) 



Figure 6-11 B 




6.4 Inequalities Involving Triangles 267 

We must prove the following three statements: 

(1) AB + BC>AC 

(2) AC +BOAB 

(3) AB + AC > BC 

(1) By hypothesis, BC > AC. Since AB > T (that is, AB is a 
positive number), we have 

AB + BC>AC 

by the Addition Property of Order (0-4). 

(2) By hypothesis, BC > AB. Since AC>O t we have 

AC + BC > AB. Why? 

(3) On opp AC choose point D so thai AD = AB. (See Figure 
6-12.) 




C Figure G-12 

Since A is between C and IX A is in the interior of L DBC. Also 
mLDBOmlABD by Theorem 65. But r»ZABD = 
mLADB by the Isosceles Triangle Theorem, and so 

mlDBOmlADB. 

"Therefore, by the Side-Comparison Theorem (applied to 
A DBC), 

DC>BC 

Since DC = DA + AC (Why?) and DA = AB, we have 

DC = AB 4- AC 

We have shown that DC > BC and that DC = AB + AC. 
Therefore, by the Substitution Property, 

AM + AC>BC 

It follows from Tneorem 6.10 that, in informal language, the short- 
est distance between two points is the length of the segment joining 
them. In formal geometry, of course, once a unit of distance is given, 
there is only one distance between two points. 

It follows from Theorems 6.8 and 6.9 that, in any one triangle, the 
greater angle lies opposite the greater side and conversely, the greater 
side lies opposite the greater angle. Let us now consider two compan- 
ion theorems involving two triangles. 



268 Inequalities In Triangles Chapter 6 

In Figure 6-13 t AB = A'W and EC = BC. What can you say 
about these triangles if m LB=mL B'? How do AC and A 'C compare? 
B a* 





Fitfirt-6-13 A C A' 

In Figure 6-14, we again have AB = AB' and BC = B'C\ but this 
time m L B > m L B\ Is the correspondence ABC < — *■ A'BC/ a con- 
gruence? I Tow do AC and A'O compare in this case? 

B B' 




A A' 

Flip j re 6-14 

In Figure 6-15, we have AB = A f B\ 
m£B<C.ni£W t Ts the correspondence ABC * — i 
ence? How do AC and A'C compare in this case? 

B' 




BC-BC 7 and 
A'BC a congru- 




A' 




c 



Figure 615 A 

On the basis of the preceding discussion, it would appear that if 
two sides of one triangle are congruent, respectively, to two sides of a 
second triangle and if the corresponding included angles are not con- 
gruent v then the sides opposite these included angles are not congru- 
ent and the side opposite the larger angle is the larger side. 

It may help you to understand this last statement by examining a 
pair of compasses. Observe that, as the angle formed by the pointers 
of the compasses gets larger, BO does the distance between the ends of 
the pointers. We make this idea formal with our next theorem. 

THEOREM 6.11 (Side-Comparison Theorem for Tivo Triangjtes) 
If two sides of one triangle are congruent, respectively , to two sides 
of a second triangle, and if the angle included by the sides of the 
first triangle is greater than the angle included by the sides of the 
second triangle, dieu the third side of the first triangle is greater 
the third side of the second triangle. 



6,4 Inequalities Involving Triangles 269 

restatement: Given A ABC and AA'B 'C w ith AB — A*ff and 
BCsrFC. If Z B> LB\ then aC>~FU. (See Figure 6-16.) 

B Hr 





C Figure 6-16 

Proof: By the Angle Construction Theorem, there is a point (J> on the 



*— > 



Oside of AB such that ZAB(<) = ZA'JJ'C", (See Figure 6-17.) Since 
A ABC > ZA'FC, P is in the interior of /.ABC, Choose point P on 

such that BF -- There are now three cases to consider. 



Cose J. P is in the exterior of A ABC, 

Case 2, P is on AC 

Qwf' 3. F is in the interior of AABC, 

We consider only Case 1 here- The proofs of Cases 2 arid 3 are left 
as exercises. 

Proof of Case h In Figure 6-17, AABP s A A 'B'U by S.A.S. Let BR 
he the midray of Z CBP intersecting AC at £. Then APBE ss ACB£ 
by S.A.S. Applying the Triangle Inequality Theorem to AAEP, we 
have AE + EF> AP. 





Figure 6-17 

But AP a A'C (Why?) and £P = EC. Why? By the Substitution 
Property, we get AE + EC > A'C, and by the Distance Betweenness 
Postulate, AC > A'C. Therefore T^C > /TTT and this completes the 
proof for Case 1. 



270 Inequalities in Triangles Chapter 6 

The converse of Theorem 6.1 1 is also true and we state it as our last 
theorem of this section. 

THEOREM §.12 (Anglo-Comparison Theorem for Two Triangles) 
If two sides of one triangle are congruent, respectively, to two sides 
of a second triangle, and if the third side of the first triangle is 
greater than the third side of the second triangle, then the angle in- 
cluded by the two sides of the first triangle is greater than the angle 
i iK- hided by the two sides of the second triangle, 

restatement: Given A ABC and AA'B'C with A~B ^ A^W and 
BC^&C, if ZC>A 7 &> then /_B> IB . (See Figure 
6-18.) 





Figure &I8 

Proof: Since m Z B an d nt Z B' are numbers, one of the following must 
hold: 

(1) mlB<mlB' 

(2) mlB=mlB' 

(3) ml.B>mLW 

Which property of numbers are we using here? 

The method of proof is to show that (I) and (2) are impossible, so (3) 
must hold* and the theorem will be proved. The problem of showing 
that (!) and (2) are impossible is left as an exercise. 

EXERCISES 6.4 

t. Given A ABC with AZ? = 12, BC = 15, and AC - 10. name the. angles 
in order of size beginning with the angle of least measure. 

2, In A $KM S mlS = 47, mlK = 85 t andml M = 48. Name the short- 
est side; the longest side. 

a Name the longest side of A ABC if (a) ml A = 44, mlB = 90; 
(b) ml A = 120, mlB = 40. 

4, Does a triangle exist with the following numbers as side lengths? Why? 
(a) Si 3, 10 (b) 5, 3, 8 (ci 5, 3, 4 

5. Given A ABC with A-B-D, if ml ABC > mlCBIX prove AC > BC 



6.4 Inequalities Involving Triangles 271 

6. Given the following figure with angle measures as marked, for each of 
the three triangles name the sides of the t dangle in order of increasing 
length 




7. In the figure for Exercise 6, which segment is the shortest? 

8. Given the following figure with angle measures as marked, prove that 
CD is the longest segment. 




9. In the figure, if the angles have the indicated measures, which segment 
is shortest? 




In Exercises 10-12, got the "best" answer you can in the sense that the 
smaller number is as large as possible and the larger number is as small as 
possible, 

10. Copy and complete: If the lengths of two sides of a triangle arc 7 and 
12, then the length of the third side must be greater than Q] and less 
than [?]. 

11. Copy and complete: If the lengths of two sides of a triangle are 6 and 9, 
then the third side must have a length less than JTJ and greater than [T|, 

12. Between what two numbers must the length of the third side of a tri- 
angle lie if two of its sides have lengths of 17 and 28? 



272 Irwqualtties in Triangles 



Chapter 6 



13, Given A-C-D, E-C-B, and with angle measures as marked, prove that 
HE > AD. 




14. Prove that the sum of the lengths of the diagonals of a convex quadri- 
lateral is less than the perimeter of the quadrilateral 



15, Given AABC with D a 
point on AC such that HD 



bisects /.ABC, prove that 
AB > AD. 



16, In the figure, FS < SA 
and PQ < QR. Prove that 

m£SPQ>m£SRQ. 




17, Prove Corollary 6,9.1. (Hint: You will need to use Corollary 6.6.1 and 
Theorem 6.9 in your proof.) 

18, Prove Corollary 6.9.2. (See Figure 6.9.) 



Given AF > PB, prove 

m L AC? > m L RCF. {Hint: 

Use Theorem 6.12.) 



Given AFQR with M 
die midpoint of TQ, if 
mlRMQ>m/ PMR, 
prove that @B > ER, 




6,4 Inequalities Involving Triangles 
D 

'C 



273 




21. Given a convex quadrilateral ABCD 
with AD = BC, if AB > DC com- 
pare m Z CAD with m Z BCA. 

22. Prove the following theorem. 



THEOREM TIa Y b,c are the side lengths of a triangle and if h a is the 
altitude corresponding to base a, then h a < h and /i a < c. 
23. chai.i.fnce problem. Prove Case 2 of Theorem 6.11. This is the Case 

whore point P is on SC, (Hint: In Figure fi-i7 for Theorem 8.11, recall 
that AABP=z A A' JVC You must show that AC > A'C.1 





24. challenge problem. Prove Case 3 of Theorem 6.11. Tins is the Case 

where point P is in the interior of A ABC. (Hint: Let BR he the bisector 
ray of ACBP, intersecting AC at E. Why is Ai^E a ACBE? Why is 
A£ + BC > A/»? Prove /££ > 5^?.) 




25. challenge problem* Complete the proof of Theorem 6. 12 by show- 
i tig tliat Cases 1 and 2 are impossible. (Hint: Use Theorem 6. 1 1 for Case 
1 and the S.A.S. Postulate for Case I.] 



CHALLENGE PROBLEM. If P is any 

point in the interior of A ABC,, prove 
that AP + PB<AC + CB. (Hint: 

Let AP intersect W at D. Apply the 
Triangle Inequality Theorem to 
AACD and to ABDF,) 




274 Inequalities in Triangles Chapter 6 

CHAPTER SUMMARY 

In this chapter we deed I with geometric inequalities involving angles 
and sides for any one triangle and also for two triangles. We defined 
GREATER THAN and LESS THAN for angles in terms of their measures 
and for segments in terms of their lengths. 

We stated some order properties for numbers which are listed below 
by name only. You should lenow the complete statement of each of these 
properties. 

0-1 POSITIVE CLOSURE PROPERTY 

6-2 TRICHOTOMY PROPERTY 

0-3 TRANSITIVE PROPERTY 

0-4 ADDITION PROPERTY 

0-5 MULTIPLICATION PROPERTY 

The key theorem concerning geometric inequalities is the EXTERIOR 
ANGLE THEOREM which states that an exterior angle of a triangle is 
greater than either of its two nonadjaeent interior angles. 

Other important theorems involving inequalities in any one triangle 
arc listed below by name only, ft is important that you be able to state these 
theorems in your own words and that you understand their proofs. 

THEOREM 6.8 (77u? Angle Comparison Theorem) 

THEOREM 6,9 (The Side-Comparison Theorem) 

THEOREM 6.10 (Tlie Triangle Inequality Theorem) 

Tile following two theorems give inequalities concerning two triangles. 

Be sure that you know the complete statement of these theorems. 

THEOREM 6.11 (The Side-Comparison Theorem for Two Triangles) 

THEOREM 6T2 (27k? Angle-Comparison Theorem for Two 
Triangles) 

We defined the DISTANCE between a point and a line not containing 
the point to be the long til of the perpendicular segment joining the point to 
the line. The distance between a line and a point on the line is defined to be 
zero. 

We proved that the hypotenuse of a right triangle is the longest side of 
the triangle and that the shortest segment from a point not on a line to the 
Hue is the segment perpendicular to the line. 



Rovltw Exercises 



275 



REVIEW EXERCISES 

In Exercises 1-7. identify the order property that Is illustrated, 

1. If m/ A>mZBandmZ.B = mZC, theu mLA>m£C. 

2. If x - y > 5 and y = 4 t then x > & 

3. If m£A> mlB, then JtwZA > |»»Zfl. 

4. If i- /- y, then % > y or * < y, 

5. If A£ < ES and BC < S£ then AB + BC < RS + ST, 

6. If i + i/ > z, then x > s — r/. 

7. If o + fe > c and a * /> < d t then d > c. 

8. In the figure, what must he true about tnAD? About mZ /JF£? AbouL 




I 



•:-: 



2) 



J? 



Refer to Figure 6-19 for Exercises 9-14. Copy each exercise and replace the 
question marks by the symbol < f =, or > to make a true statement. 



9, h Q] tl 

10. ag]c|T]e 

11. cQd 

12. c[T|/ 

13. fl|?J/ 




14. b^]d[T\f\2\c\2XQ 

15. Which theorem do the markings 
on the figure contradict? 



Figure 6-19 

16. In the figure, which angle is 
the largest? The smallest? 





276 Inequalities in Triangles 

17, In the figure, which side is the longest? The shortest? 



Chapter 6 




18. In the figure, if the angles have the indicated measures, which segment 
is shortest? What theorem did you use to decide? 




19, In the figure, if the ingjtea have the indicated measures, arrange the seg- 
ments PQ, QR, rT5, ?7S, PS, undTTi in order of increasing length. 




20, Copy and complete th«- fi blowing statement, putting the largest number 
in (a) and the smallest number in (b) which will make a true statement 

If the lengths of two sides of a triangle are 8 and 15. then the 
length of the third side must be greater than (a) (?] and less 

than (b) [T]- 



Review Exercises 277 



Exercises 21-26 refer to Figure 6-20 showing A ABC with A-D-C. B-E-Q 
B-F-D, and A-F-E, You should be able to defend your answers using the- 
orems that you know. 




Figure 6-20 

21. Name five angles whose measures arc less than m L 10. 

22. Name two angles whose measures are less than mZ.5. 

23. Name four angles whose measures are greater than mL 1. 

24. L 7 is an exterior uugje of which two triangles? 

25. Is /_ 6 an exterior angle of ABFE? 

26. Is ml& > mZ2? Explain why. 

27. In the figure, if BC is the longest side and 7%\ is the shortest side, prove 
that m/,A > m£C, (Hint: Draw GA and use the Angle-Comparison 

Theorem.) 




28. Given: SK = KM 
ftere KM > KJ 

[Hint; Use the Kxterior Angle Theorem, the Isosceles Triangle Theorem, 
and the Side-Comparison Theorem.) 




278 Inequalities in Triangles 

29. For A ABC prove that AH - WO < AC, 



Chapter 6 




30. If, In AAjBC, BC = AC, prove that AC > JAB. 

31. Given th e fig ure with angle and side measures as marked, name the 
sides SR t RQ t and QPin order of increasing length. State a theorem that 

justifies your conclusions. 




32. Given the figure with sides us marked, copy and insert < or > in (a), 
(l>), and (c) so that each is a true statement. 

(a) roZAJTJmZBEC 

(b) mZDEC[?]mZBEC 

(c) ittZJi 0mZDEC 




33. Slate theorems that justify your conclusions in Exercise 32, 



R*vl*w exercises 279 



34. Given: &ABC with median AM 
mLAMB = 70 

Prove; mABymlC 




CMS 

35, If ASOf is a right triangle with S-M-C, prove that KC > KM 

K 




36. challenge problem. If A, B, C, D are four distinct points in space 
and if no three of these points lie on a line, prove that 

AB + EC + CD> DA. 

37. challenge PROBLEM. Prove that in any triangle there are two sides of 
lengths r and s such that 

±<t<2. 




Dat;kl Phnctltm/Phuto Ke$eurclwr,\ 



Parallelism 



7.1 INTRODUCTION 

We are now ready to consider one of the most basic ideas in geom- 
etry, the idea of parallel lines. What do yon think of when someone says 
"parallel lines?" Perhaps the lines which separate lanes on a straight 
running track or the cracks between the boards in a floor, or the strings 
on a violin? What are some of the properties of parallel lines? 

If two ships are sailing parallel courses close to one another as sug- 
gested in Figure 7-1 , what is the relationship of the angles marked in 
the figure? 





Figure 7-1 



282 Parallelism Chapter 7 

The congruence of these angles is an example of an Idea from in- 
formal geometry that suggests a property of parallel lines in formal 
geometry. 

The set of rails on a railroad track must fit wheels which arc fixed so 
that the distance between them cannot change as they roll down the 
track. This suggests another property of parallel lines in formal georn- 
eLry, the property of being the same distance apart everywhere. 

What arc the basic properties of parallel lines? Can you think of one 
or more properties such that if lines have these properties they should 
then be considered to be parallel lines? It would be natural to use such 
basic properties in deciding on a definition for parallel lines. 

ITie most basic properties of points, lines, and planes are the Inci- 
dence Properties which were discussed in Chapter L Two different 
lines either intersect or they do not, IF they do not intersect and are 
copianar, we call them parallel lines. If they do not intersect and are 
not copianar, we call them skew lines. We know that two distinct in- 
tersecting lines determine a plane. It follows that, if two lines are not 
copianar, then they must he nonintersecting lines and hence are skew 
lines. We shall find it convenient to consider a line as parallel to itself. 
We begin our formal treatment with several definitions based on the 
foregoing ideas. 



7.2 DEFINITIONS 



Definition 7.1 Two distinct lines which are copianar and 
nonintersecting are parallel lines, and each is said to be paral- 
lel to the other. Also, a line is parallel to itself. The lines in a 
set of lines are said to be parallel lines if each two in the set 

are parallel. 



Definition 7.2 TWo lines which do not lie in the same plane 
are called skew lines. 



\ citation. If p and q are lines, then p | q means that p is parallel to (f, 

and pX° means that p is not parallel to q. 

Note that every pair of distinct parallel lines arc copianar, that 
every pair of distinct intersecting lines are copianar, and that every 
pair of skew lines are noncoplanar. If p and q are lines, there are four 

distinct possibilities: 



7.2 Definitions 283 



1. p and q are noncoplanar, in which case they axe skt^u . 

2. p and q are coplanar but not parallel, in which case they inter- 
sect in exactly one point and there is exactly one plane which 
contains them. 

3. p and q arc distinct parallel lines, in which case they do not in- 
tersect and there is exactly one plane which contains them. 

4. p and q are nondistJnct parallel lines, in which case p and q are 
the same line and there are infinitely many different planes con- 
taining p and q t 

EXERCISES 7.2 

Exercises 1-5 pertain to the rectangular box suggested in Figure 7*2. 

E 




Figure 7-£ 

1. Using A, H, C, , . ■ , designate two distinct edges which lie ad parallel 
lines. 

2. Using A, B, C, . , , , designate two distinct edges which lie on intersect- 
ing lines. 

3. Using A, J3 T C, . , , , designate two distinct edges which lie on skew lines. 
4 How many edges of the box lie on lines which are parallel to AB? 

5. How many edges of the box lie on lines which are skew to TiB? 

6. If p and q are skew lines, why is there no plane which contains both? 

7. If p is a line in a plane a t how many different lines in a are parallel top? 

8. If p is a line in a plane a t how many different lines in a are perpendicular 
top? 

0. If p is a line in a plane a, how many different lines in a are skew to p? 

10, Prove that two distinct parallel lines detenninc exactly one plane. 

11, Given that m is a line and P is a point on tit, prove that there is one and 
only one line through P and parallel to RL 

12, Let m be a line in a plane a and P a point in a but not on wi. Use your 
knowledge regarding the existence of perpendicular lines and the Ex- 
terior Angle Theorem for triangles to prove that there is at least one 
line through P and parallel to m. 



284 Parallelism Chapter 7 

7.3 EXISTENCE OF PARALLEL LINES 

Answering questions regarding existence is an important part of 
building a formal geometry. How do we know that there are such 
things as poinLs, lines, and planes? We know that they exist because 
of the postulates and the theorems of Chapter I on incidence relations. 
Given two distinct points A and B on a line l t how do we know that 
there is a point Con/ such that A-C-B and AC - CB? We know it 
liecause we can prove it! Our proof depends in a very essential way on 

the Ruler Postulate, Given a line A B in a plane a, how do we know that 

there exists a ray AC in a such that m/.RAC = 30? We know it be- 
cause we can prove itl Of course, the proof is easy once we adopt die 
Protractor Postulate- 
Given a line I and a point P not on I, how do we know that there 
exists at least one line through P and parallel to J? It is true in some ge- 
ometries that there are no parallel lines. But in our geometry t the ge- 
ometry that we inherited from Euclid, there are parallel lines, and 
furthermore, we can prove it. In fact, you were asked to prove it in 
Exercises 11 and 12 of Exercises 7*2, These exercises arc combined in 
the next theorem. 

THEOREM 7.1 (Existence of Parallel Lines Theorem) If I is a 
line and P is a point, then there is at least one line through P and 
parallel to I. If P is on I, Hi ere is exactly one line through P and paral- 
lel to i 

Proof: There are two cases to consider: (1) P is on I and (2) P is not on I 

Case 1. We suppose first that P is on if. Since I is parallel to I, it follows 
that there is a line through P and parallel to L If m is any line different 
from I and through P t then it intersects I in exactly one point and hence 
is not parallel to L Therefore there is one and only one line through P 
and parallel to I. 

Case 2. Suppose next that P is not on I (See Figure 7-3.) Then P and / 
determine a plane. Why? Gall It a. 



4 ll 

Figure 7-3 



7.3 Existence of Parallel Lines 

In a there is a unique line m through P and perpendicular to I (see 
Theorem 6.7) and a unique line n through P and perpendicular to m. 
We shall prove that n and / are parallel. 

Suppose, contrary to what we want to prove, that n and / are not 
parallel. Then m, 11, and / are distinct intersecting lines forming a tri- 
angle with an exterior angle and a nonadjaccnt interior angle both of 
which are right angles. (See Figure 7-4.) But this is impossible in view 
of the Exterior Angle Theorem. It follows that n and I arc parallel. 
Therefore there is at least one Hue through Panel parallel to I This com- 
pletes the proof. 



285 




Figure 7-4 

Along with questions regarding existence in mathematics there are 
sometimes questions regarding uniqueness. Theorem 7.1 settles the 
matter of existence of parallel lines, but it docs not settle the matter of 
uniqueness. If / is a line and .Pis a point on /, we know that there is one 
and only one line through F and parallel to L If / is a line and P is a 
point nof on I, we do not know yet that there is one and only one line 
through P and parallel to L For about 2<XX) years following the time of 
Euclid, mathematicians tried to prove that, given a line and a point not 
on the line, there is a unique line through the given point and parallel 
to the given line. Finally two mathematicians, a Russian named Nikolai 
Ivanovitch Lobachevsky (1793-1856) and a Hungarian named Janos 
Bolyai (1802-1860), proved independently that it is impossible using 
only the postulates of Euclid (other than his parallel postulate) to prove 
the uniqueness of parallels. Since we want uniqueness of parallels, we 
follow hi the footsteps of Euclid and adopt a "parallel postulate," We 
defer the statement of lliis postulate, however, to Section 7,8. 

In Sections 7.4 and 7.5 we introduce the concept of a transversal 
and develop some theorems whose proofs do not depend on the Parallel 
Postulate. These arc theorems that belong both to the ordinary geom- 
etry of Euclid and to the non-Euclidean geometry of Lobachevsky and 
Bolyai in which the Euclidean Parallel Postulate is replaced by a postu- 
late which says that parallels are not unique. More specifically, the 
Bolyai-Lobachevsky Postulate states that if Ms any line and P is any 
point not on 1, dien there are at least two distinct lines through P and 
parallel to I 



28« 



Parallelism 



Chapter 7 



7.4 TRANSVERSALS AND ASSOCIATED ANGLES 

Let p, (/, and f be three distinct lines in a plane a. The line t may in- 
tersect both p and q or it may not If t intersects both p and q, then it 
may intersect them in different points or it may intersect them in the 
same point In Figure 7-5, p and q are intersecting lines and t intersects 
p and q in different points. In Figure 7-G, p and q are parallel lines and 
t intersects p and q in different points. In Figure 7-7, t intersects both 
p and q in the same point. In Figure 7-8, t intersects q but does not 
intersect p. 





Figure 745 




P + 



< + 



7 



Figure 7-7 



Figure 7-S 



In situations like those in Figures 7-5 and 7-6 we say that f is a 
trunsc&rsal of p and q. In each of these figures t intersects the union of 
p and 9 in a set consisting of two distinct points. On the other hand, in 
Figures 7-7 and 7-8, t docs not intersect the union of p and q in a set 
consisting of two distinct points. As we said, in Figure 7-5, / is a trans- 
versal of p and q. In Figure 7-5 q is a transversal of p and f, and p is a 
transversal of q and f. We arc now ready for the following formal 
definition. 



7.4 Transversals and Associated Angles 287 



Definition 7.3 A transversal of two distinct coplanar lines is 
a line which intersects their union in exactly two distinct 
points. 



In Figure 7-9, A s B t C, D are four of the vertices of a rectangular 



box. 




li Figure 7-9 

Note that 

AB, BC, CD are three distinct lines, 

BC intersects the union of .4 B and CD in two distinct 
points. Name them. 

is not a transversal of AB and CD. Why not? 



Is it true that if a line is a transversal of two other lines, then the 
three lines arc distinct eoplanar lines? Give a reason for your answer, 

A transversal of two lines forms with these lines eight distinct an- 
gles. For convenience we gjve names to certain pairs of these angles. 




Figure 7-10 



In Figure 7-10, * is a transversal of p and c/; angles 1, 2 t 3, 4 are the 
angles formed by p and i; angles 5, 6, 7, 8 are the angles formed by q 
and t. Angles 1, 4, 6, 7 are called ulterior angles and angles 2, 3, 5, 8 
are called exterior angles , 



2*K 



taniliiin 



er7 



Angles 1 and 6 are one pair of consecutive interior angles. Figure 

7-1 1 shows these angles with several points labeled. Notice that Z 1 is 

the union of BA and ED and that Z 1 is LABD. Express Z 6 in terms 
of rays and write a name for it involving names of points. 




Figure 7-11 9 



Notice that Z 1 and Z 6 have a segment in common and that their 
interiors are on the same side of the transversal. Name the segment that 
is die intersection of these two angles. 

Lines p and q t which we are considering in connection with a trans- 
versal f, might be parallel or they might not be. if p and q are not paral- 
lel, then they intersect in a point. This point might be on the opposite 
side of the transversal from the interiors of a pair of consecutive in- 
terior angles, as it would be for Z 1 and Z 6 in Figure 7*11, or it might 
be on the same side, as it is for the pair of consecutive interior angles 
4 and 7 in Figure 7-12. Note that although the intersection of Z 1 and 
Z 6 is a segment, the intersection of Z4 and Z 7 is the union of a seg- 
ment ami a set consisting of a single point. Name that segment and that 
point. Thus the intersection of two consecutive interior angles may 
be a segment or it may be the union of a segment and a set consisting 
of a single point. 




Figure 7-12 

We shall give a formal definition of consecutive interior angles after 
we have introduced two other phrases for angles associated with u pair 
of Uiies and a transversal. 



7.4 Transversals and Associated Angles 289 



In Figure 7-13, Z 1 and L 7 are interior angles but not consecutive 
interior angles and not adjacent angles. We call them alternate interior 
angles. The intersection of Z 1 and Z 7 is a segment and their interiors 
are on opposite sides of the transversal. 




Figure 7-13 

Another pair of alternate interior angles are Z 4 and Z 6, (See Fig- 
ure 7-14.) Their intersection is a segment and their interiors lie on op- 
posite sides of the transversal. 




Figure 7-14 

In Figure 7-15, Z I and Z 6 are consecutive interior angles. Angles 

1 and 2 are adjacent angles. Angles J. and 7 are alternate interior angles. 
This brings us to Z 1 and £5. 




Figure 7-15 



290 



Parallelism 



Chapter 7 



Angle 1 is an interior angle and Z 5 is not. Their interiors lie on the 
same side of the transversa]. In Figure 7-16, these angles are shown 
with several points labeled. Examine the intersection of Z 1 and £5. 




Figure 7-ia 

Is the intersection of these angles a ray s the ray BP? Observe Lhat the 
intersection of die interiors of Z 1 and Z 5 is the interior of Z 5. Angles 
1 and 5 are called corresponding angles. 

Another pair of corresponding angles associated with the lines p 
and q and their transversal t are Z 4 and Z 8 as shown in Figure 7-17. 
The intersection of these angles is the union of a ray and a set consist- 
ing of a single point. The interiors of these angles lie on the same side 
of the transversal. Although neither interior contains the other, they do 
intersect. 




Figure 7-17 

Thus far in this section we have introduced alternate interior an- 
gles, consecutive interior angles, and corresponding angles in connec- 
tion with two coplanar lines and a transversal. Actually, it is not neces- 
sary to refer to these lines and the transversal in describing these angles. 
Indeed, no such reference is made in the following definitions. 

The phrase "a segment and a point*' appears in the following defini- 
tions. We accept these phrases as a short way of saying "die union of 
two sets of points, one of them a segment and the other a set which 
consists of a single point/' We accept "a my and a point** to mean "the 
union of two sets, one of them a ray and the other a set consisting of a 
single point." 



7.4 Transversals and Associated Angles 291 



Definition 7.4 Two coplanar angles are alternate interior 
angles if their intersection is a segment and if their interiors 
do not intersect. 

Definition 7.5 Two coplanar angles are consecutive interior 
angles if their intersection is a segment, or a segment and a 

point, and if their interiors intersect. 

Definition 7.6 Two coplanar angles are corresponding an- 
sxlcs if their intersection is a ray, or a ray and a point, and if 
their interiors intersect. 



EXERCISES 7,1 

1, How do yon know that parallel lines exist in Euclidean geometry? Is 
your reason a postulate or a theorem or u definition? 

& If p is a line, if q is a line, and if p is parallel to q, is it possible that the 
intersection of p and q is a set which contains more than one point? 
Explain. 

3. If p is a line, if q f§ a line, and if the intersection of p and q is the null 
set, is it possible that p and q are not parallel? Explain. 

4. If p is a line, if q is u line, and if p is parallel to q, is it possible that there 
is no plane containing the union of p and q? Explain, 

■ Figure 7-18 sh o ws t wo coplanar lines p a nd q a nd a transversal t with several 
points labeled. Copy Exercises 5- 10 and replace the question marks with 
one word,, two words, or three capital letters so that the resulting statement 
is a true sentence concerning Kijjyre 7-18. 




Figure 



5* LBDC and ZQ arc corresponding angles, 
fl, / BDC and Z[7] are alternate interior angles. 
7. L BDC and Z |T| arc consecutive interior angles, 
a L FDC and LAQD are \?J angles. 
9. / HDF and Z HOP. are CD angles. 
10, Z FDC and / ECD arc [T| angles. 



292 



Parallelism 



Chapter 7 



Figure 7- 19 shows two copbnur lines p and q and a transversal t. Eight an- 
gles are labeled. Copy Exercises 1 1 -20 and replace the question marks with 
one word or two words or a letter or a number so as to make a true statement 
about Figure 7-19, 




BrgHK 7-19 

11. a and C are vertical [?]. 
1& a and d are \T\ angles. 

13. a and u are [?] angles, 

14. c and c are [7| angles. 

15. c and [7] are corresponding angles. 
16\ x and [f| are corresponding angles, 

17. d and [?] are alternate interior angles. 

18. There are JT| pairs of alternate interior angles among the eight angles. 

19. Thereare {7] pairs of consecutive interior angles among the eight angles. 

20. There are \T\ pairs of corresponding angles among the eight angles. 

<■ — ■» < — * «. — t 
Figiue 7-20 shows three iioncollineur points A, B f C, the lines AB, HC, CA, 

and 12 angles formed by than, Copy Exercises 21-23 and replace the ques- 
tion marks so as to make a true .statement about the figure. 




Figure 7-20 

21. / 1 and ]7] are corresponding angles; also / 1 and [Tj are corresponding 
angles. 

22. Z 12 and \T\ are alternate i ntcrior angles; also L 12 and [?] are alternate 
interior angles. 



7,4 Transversals and Associated Angles 293 

23. _ 1 2 and IT| are consecutive interior angles; also / 12 aud [T] are con- 
secutive interior angles, 

Figure 7-21 shows two distinct parallel lines, a transversal, and eight associ- 
ated angles. Copy Exercises 24-28 and replace the question marks so as to 
make a true statement about the figure. 



"4- 




Figure 7-21 

24, Angles 1 and 5 arc corresponding angles; their intersection is Q]; I heir 
interiors lie on [7] of L 

25, Angles 3 and 5 are consecutive interior angles; their intersection is [Tj: 
their interiors lie on [7] of t, 

26, Angles 3 and 6 are alternate interior angles; their intersection is Q]; their 
interiors lie on [7] of f , 

27* Angles 6 and 7 are vertical angles; their union is the [tj of a pair of lines; 

their intersection is Q]; the intersection of their interiors is [7], 
28. Angles 5 and ft are a linear pair of angles; their union is the union of a 

line and a [T|; their intersection is JT]; the union of these angles and their 

interiors is the union of a halfplane and Q], 

20, The figure shows A ABC and two rays in the same plane such that B is 

between A and E r and BD is between liE and EC. Considering only 
angles that can be named in terms of three points labeled in the 

figure, identify all pairs of alternate interior angles, 




30. Under the same conditions as those in Exercise 29„ identify ail pairs of 
consecutive interior angles, 

31. Under the same conditions as those in Kxercise 29, identify all pairs of 
corresponding angles. 



294 ParalJefism 



Chapter 7 



Figure 7-22 shows a quadrilateral and one of its diagonals. In Exercises 
32-35, copy and complete each statement so that it will be true. 




32. / ACH and Z CAD are alternate interior angles detennmed by trans- 
versal aS and lines \f\ and [7]. 

33. Z BAC and / DC A are alternate interior angles determined by trans- 
versal |TJ and lines \T\ and {T}, 

34 Z ABC and Z BCD arc \T} angles determined bv transversal jT] and lines 
(U and Q] . 

35. Z BAD and Z arc consecutive interior angles detennined by trans- 
versal [T\ and lines and [7). 

Figure 7-23 is a plane figure with five angles labeled, including two pairs 
of vertical angles. Let m Z a m 65 and mZa = m / e. In Exercises 36-38 t 
find the measure of the given angle 



36. Li 

37. Zu 

38. Ze 



Figure 7-23 




Figure 7-24 is a plane figure showing two lines and a transversal. Refer to 
tills figure for Exercises 39-42. 




Figure 7-24 



7.5 Some Parallel Line Theor&m*, 295 

39. Name 

(a) the pairs of alternate interior angles; 

(b) tlie pairs of consecutive interior angles; 

(c) the pairs of corresponding angles. 

40. Prove: If one pair of alternate interior angles are congruent, then 

(a) the other pair of alternate interior angles are congruent; 

(b) each pair of consecutive interior angles are supplementary: 

(c) each pair of corresponding angles arc congruent 

41. Prove: If one pair of consecutive interior angles are supple mentary, 
then 

(a) the other pair of consecutive interior angles are supplementary; 

(b) each pair of alternate interior angles arc congruent; 

(c) each pair of corresponding angles are eongruent 

42. Prove: If one pair of corresponding angles are congruent, then 

(a) each pair of corresponding angles are congruent; 

(b) each pair of alternate interior angles arc congruent; 

(c) each pair of consecutive interior angles are supplementary, 

7.5 SOME PARALLEL LINE THEOREMS 

This section contains several theorems that are sometimes helpful 
in proving that two lines are parallel. 

THEOREM 7,2 Let two distinct coplanar lines and a transversal 

be given. If the transversal is perpendicular to both lines, then the 
lines are parallel, 

Proof: Let a and b be distinct coplanar lines and let t be a transversal 
of them. (See Figure 7-25.) We wish to prove that if a and h are per- 
pendicular to t t then a and b are parallel. Suppose, contrary to this as- 
sertion and as suggested by the figure, that a _L t. that feif, and that 
a is not parallel to b. Then a and b intersect in some point P forming 
APVQ. This triangle has Z 1 as an exterior angle and Z 2 as a nonadja- 
cent interior angle. By hypothesis, both of these angles arc right angles, 
so &F¥Q is a triangle with one exterior angle congruent to a nonadja- 
cent interior angle. 




296 Parallelism Chapter 7 

According to the Exterior Angle Theorem, however, the measure 
of an exterior angle of a triangle is greater than the measure of cither 
nonadjacent interior angle- 1 truce we have two angles which are con- 
gruent by one line of reasoning and which are not congruent by an- 
other line of reasoning. We arrived at this contradiction after we had 
supposed that die lines a and h are not parallel Since we cannot have a 
contradiction in our system, we have proved that lines a and h cannot 
lie not parallel. Hence they are parallel This completes the proof of 
the theorem. 

We pause briefly to comment on this proof. It is an example of an 
indirect proof, The theorems in our formal geometry arc true state- 
ments. Their truth rests on a foundation of postulates and definitions. 
In proving theorems, that is, in establishing their truth, we may use 
definitions, postulates, and theorems proved previously. 

flow did we prove Theorem 7i2? Consider the given situation in- 
volving lines a T h r and t. Lines a and h are distinct and coplanar, and / 
Is perpendicular to both of them. We do not know that a and h are par- 
allel nor do we know that a and b arc not parallel But we do know 
from our definitions that a and b are either parallel or they are not par- 
allel Since one of these possibilities leads to a contradiction, we con- 
clude that the other possibility is the valid conclusion. 

It may be helpful to see this reasoning in skeleton form using sym- 
bols. We start with a situation in which a and b are coplanar lines and t 
is a transversal of them. Let P and Q stand for statements as follows: 

P: t ±a and t 1 b. 
Q. tffc 

Then our theorem and proof arc essentially as follows. 
THEOREM If P, then 0. 

Proof: It is given that P is true. Either Q is true or Q is false. If QU 

false, then it follows that Ph false, This contradicts the hypothesis that 
P is true, Therefore Q is true, and the proof is complete. 

If P denotes a statement, it is convenient to denote the opposite 
statement by not-H If P is a true statement, then not-F is a false state- 
ment If P is a false statement, then not-P is a true statement. If a state- 
ment has the form "If P, then Q" then there is always a related state- 
ment of the forrn "If not-Q* men not-JV' This related statement is 
called the contrapositivc of the given statement. If the contrapositive 
of a statement is true, then the statement is true. For, if not-Q implies 
not-P, then the only way it is possible for J* to be true is for {) also to be 



7.5 Some Parallel Line Theorems 297 

true. In other words, if not-Q implies not-P, then P implies Q. This 
means that we can prove a theorem by proving its contrapositive. If it 
is easier to prove the eontrapositive, then this is what we should do. A 
proof using the contrapositive is one form of indirect proof. 
For example, suppose we wish to prove 

if a 2 =f=b 2 s then a^=h. 

An easy way to prove this is to prove its conlrapositive 

if a = h, then a 2 = h 2 . 

Tlie proofs of the following three theorems are assigned as exercises. 

THEOREM 7,3 {Alternate Interior Angle Theorem) If two al- 
ternate interior angles determined by two distinct coplanar lines 

and a transversal are congruent then the lines are parallel. 

THEOREM 7.4 (Corresponding Angle Theorem) If two corre- 
sponding angles determined by two distinct coplanar lines and a 
transversal are congruent, men the lines are parallel. 

THEOREM 7.5 [Comecutwe Interior Angle Theorem) If two 
consecutive interior angles determined by two distinct coplanar 
lines and a transversal are supplementary, then the lines are parallel 

EXERCTSES 7.5 

Figure 7-26 shows u transversal of Iwo distinct coplanar lines and eight as- 
sociated angles, In Exercises 1—4, measures of two of the angles are given. 
Explain why m must lie parallel to n. 



m + 



««- 




Figure 7-26 

1. m/.a = 105, m£e = 105 3. mLc = 75, mLv = 105 

2. mid = 105, mZf= 105 4. m t a = 75, m L f = 75 



29S 



Parallelism 



Chapter 7 



Figure 7-27 shows five coplanar segments and several associated angles. In 
Exercises 5-9, state which lines must be parallel on the basis of the given 
information. 



5. Z«s Iv 

6. Zpss lq 

7. ml ADC + mlDCB = 180 

8. ml ADC + ml BAD = ISO 

9. ml DAB + ml ABC = 180 



Figure 7-28 shows five distinct coplanar points A, B, Q D, E. Points A, B, C 
are coHinear, AD 1 Tkff, and CE _L AC In Exercises 10-14, state whether 
this given information implies the Stated conclusion. 



10. A3 | AC 

11. A~$ | M 

12. a6 if ci 




13. AD | EC 

14. XBl\ BE 




In Exercises 15-20, the figures show three coplanar lines. The measures of 
three angles, expressed in terms of a number x, are given in the figures- Find 
x and determine whether m is parallel to n. Give a reason for your answer. 



15. 



16. 




7,5 Some Parallel Una Thtortms 299 




18. 



m4- 



"4- 




2x + 5 




™«- 



lOO + i/ 80 + * 




*4- 



100 + 21 



21. Given: A plane figure with AB = CD, AD = BC. 
Trove: AD \ B&. XS jj DC 




SOD 



Parallelism 



Chapter 7 



22. Given: A figure with A-D-B, A-E-Q 
AD = A£ AB = AC. 
LACBz* IADE 

Prove: DE\BC 



23. Given: A figure in which A~C and SD bisect each other at M, 
Prove: A~B \ S3, A/5 || S3 





2? C 

24. Gitcn. A figure with A-R-B, B-P-C, C-Q-A. 

AQ -QC= RP 
CF = PBzzQR 
RR = RA = PQ 

Prove: m/_A + m£B + m I C = 180 




25. Git en. A plane figure, with I A =z IB, AD = BC, A£ = EB t 
DF=m 

Prove: EF 1 W, AB J EF, DC || A~S 

U Lil £ III .C 



A £ rt 

26. Prove Theorem 7.3. 

27. Prove Theorem 7,4. 

28. Prove Theorem 7.5. 

29. Write the contraposiuve of Theorem 7,3. Is this coiitrapositive a true 
statement? 

30. Write the contrapositive of Theorem 7.4. Is this contrapositive a true 
statement? 

31. Write the contrapositive of Theorem 7.5. Is this contrapositive a true 
statement? 



7.6 The Parallel Postulate 301 



7.6 THE PARALLEL POSTULATE AND SOME THEOREMS 

Theorem 7. 1 states that through a given point not on a given line 
there is at least one line parallel to the given line. As we pointed out 
following the proof of that theorem, it is impossible, using only Pos- 
tulates 1-25, to prove that this parallel line is unique. Since we 
want it to be unique, we adopt the following Parallel Postulate. As we 
stated in Section 7.3 it is this postulate which makes our geometry, 
Euclidean geometry, different from that of Lobachevsky and Bolyai. 

POSTULATE 26 (Parallel Postulate) There is at most one line 
parallel to a gjven line and containing a given point not on the given 
line. 

This postulate and Theorem 7.1 tell us that if a line and a point not 
on it are given t then there is exactly one line through the given point 
and parallel to the given line. Furthermore, as we said in Section 7.3, 
we know that if a line and a point on it are given, then there is exactly 
one line through the given point and parallel to the given line, namely 
the given line itself. Hence, in every case, given a line and a point, 
there is exactly one line through the given point and parallel to the 
given line* 

In Section 7.5 there are some parallel line theorems, actually the- 
orems stating conditions that imply that lines arc parallel. These the- 
orems are useful in proving lines parallel. In this section we have the 
converses of several of these theorems, These converses were deferred 
until now because the Parallel Postulate is essential for their proof. 

If a theorem is of the form "If P, then Q," then its converse is the 
associated statement "If Q, then ?" It should be clear that the con- 
verses of some true statements are not true statements. For example, 
'Tf a number is greater than 100, then it is greater than 6" is a true 
statement, whereas its converse, c *If a number is greater than 6, then it 
is greater than 100," is certainly a false statement The converses of 
some theorems are theorems: the converses of other theorems are not 
theorems. We shall prove that the converses of Theorems 7.3, 7.4, and 
7.5 are theorems. 

Note first, however, that the converse of Theorem 7.2 is not true. 
To see this, suppose m and n are distinct parallel lines and that f is a 
transversal of them which makes an augle of measure 30 with m. Then, 
according to Theorem 7.7 given later in this section, it also makes an 
angle of measure 30 with line n. Thus wc see that lines m and n may be 
parallel even though a transversal is not perpendicular to both of them. 

We proceed now to the converses of the other theorems in Section 
7.5. 



302 Parallelism Chapter 7 

THEOREM 7.6 (Converse of Alternate Interior Angle Tlieorem) 
Tf two distinct lines are parallel then any two alternate interior 
angles determined by a transversal of the lines are congruent. 

Proof: (See Figure 7-29.) Let m and n be distinct parallel lines. Let t 
be a transversal that intersects them in P and Q, respectively. Let R 
and S be points on m and n, respectively, such that R and S are on op- 
posite sides of t. Wc shall prove that LRPQ == LSQT. 



t 

Figure 7-W 

It follows from the Angle Construction Theorem that there is a ray 

~PR\ with W on the tf-side of U such that LR'PQ as Z 5QK Since 
Z RTQ and Z SpP are congruent alternate interior angles, it follows 

from Theorem 7.3 that lines AT and n are parallel. Therefore AT and 
m are lines through P and parallel to n. From the Parallel Postulate, 
however, it follows that there is only one line through F and parallel 

to n. Therefore iiT and m are the same line. Then Z R'PQ and Z RPQ 
are the same angle and it follows that Z RPQ m Z SQP. 

THEOREM 7, 7 (Converse of Corresponding Angle Theorem) Tf 

two distinct lines arc parallel, then any two corresponding angles 
determined by a transversal of the hues are congruent. 

Proof; Assigned as an exercise. 

THEOREM 7,8 (Converse of Consecutive Interior Angle The- 
orem) If two distinct lines are parallel, then any two consecutive 
interior angles determined by a transversal of die lines are 
supplementary. 

Proof: Assigned as an exercise. 

As we noted above, the converse of Theorem 7.2 is not true. It 
should be noted, however, that a theorem similar to Theorem 7.2 does 
have a true converse. This is Theorem 7.9. Its converse is Theorem 
7.10. Tlieorem 7.9 follows immediately from Theorem 73. Theorem 



7.6 The Parallel Postulate 303 

7.2 begins with a "situation statement" (Let two distinct coplanar lines 
and a transversal he given) followed by an "If P, then Q'* type of .state- 
ment In Theorem 7.9 some of the P has been put into the "situation/' 
but the meaning of the two sentences taken logcdier is the same as in 
Theorem 7.2. Theorem 7.10 follows immediately from Theorem 7.7. 

THEOREM 7.9 I^et a and h be two distinct coplanar lines, and 
let t be a transversal of them that is perpendicular to a. If t is per- 
pendicular to b, the lines a and b are parallel. 

THEOREM 7.10 Let a and h l>e two distinct coplanar lines, and 
let f he a transversal of them that is perpendicular to a. If a and b 
are parallel, then f is perpendicular to h. 

The next theorem provides anodicr useful method for proving lines 
parallel, 

TTTEOR EM 7,11 Two coplanar lines parallel to the same line are 
parallel to each other. 

Proof: Let p and q be coplanar lines each parallel to a line r. If p — q t 
then p is certainly parallel to q. Suppose, then, that p and q are distinct 
lines. There are two possibilities as indicated in Figure 7-30. Either p 
and q are parallel lines or else they have exactly one point T say P, in 



14- 




P'oeelbUity (») Powubility fb) 

Figure 7-30 

common. If they have exactly one point P in common, then there arc 
two distinct lines through P and parallel to r. Since this contradicts the 
Parallel Postulate, it follows that possibility (h) in Figure 7-30 is impos- 
sible, and therefore p is parallel to q. 

If we consider only lines lying in a given plane, we see that the re- 
lationship of parallelism for lines is an equivalence relation. That is.. It 
is reflexive, symmetric, and transitive, (1) It is reflexive since every line 
is parallel to itself. (2) It is symmetric since if line p is parallel to line </, 
then line q is parallel to line p, (3) To show that it is transitive, suppose 
that line p is parallel to line r and that line r is parallel to line q. Then, 
since parallelism in a plane is symmetric, it follows that line q is par- 
allel to hue r. Then p and q are both parallel to line r and it follows from 



304 



Parallelism 



Chapter 7 



Theorem 7,11 that p is parallel to q. Later we shall see that parallelism 
for lines is an equivalence relation even without the restriction that the 
lines all lie in one plane. 

THEOREM 7A2 Let three distinct eoplanar lines with two of 
them parallel be given. If the third line intersects one of the two 
parallel lines, then it intersects the other also, 

Proof: Assigned as an exercise. 

THEOREM 7.13 Let two sets % and 3 of parallel lines in a plane 
q he given, (This means that every two lines in § are parallel and 
that every two lines in 3 are parallel.) If one line in § is perpendic- 
ular to one line in 3 , then every line in § is perpendicular to every 
line in 3. 

Proof: Let s be a line in S and let t be a line in 3 such that s is per- 
pendicular to t. Let u be any line in g and let v be any line in 3 . We 
want to prove that u is perpendicular to c. Suppose, first, that s and u 
are distinct lines and that r and i" are distinct lines as indicated in Fig- 
ure 7-31. Since * intersects t and i is parallel to t% it foDows from The- 
orem 7-12 that 8 intersects v and therefore s is a transversal of the par- 
allel lines t and o. Then it follows from Theorem 7.10 mat s is 
perpendicular to v. Similarly, it follows from Theorems 7.12 and 7.10 
thai c is a transversal of the parallel lines * and u and that v is perpen- 
dicular to u. This completes the proof for the case in which s and u are 
distinct lines and t and are distinct lines. 



i 








► 


8 


- 








u« 








► 



Figure 7-31 

Suppose next that s = u and that s is perpendicular to t. Then it 
follows as above (really the halfway point in the reasoning of the pre- 
ceding paragraph) that s is perpendicular to v. 

Tlie ease in which s and u are distinct and t = c is assigned as an 
exercise. 

The case in which s = u and I = v requires no proof since the 
hypothesis and the conclusion are the same in this instance. 



7.6 The Parallel Postulate 



305 



EXERCISES 7,6 

1. Use Theorem 7,6 to prove Theorem 7 J. 

2. Use Theorem 7.6 to prove Theorem 7.8. 

3. Prove Theorem 7.7 without using Theorem 7.6, (This proof will be sim- 
ilar to I he one for Theorem 7.6 that uses the Parallel Postulate.) 

4 Prove Theorem 7.8. (This proof will be similar to the one for Theorem 
7.6 that uses the Parallel Postulate.) 

5. Explain (he following statement; Of the three theorems, 7,fi, 7,7, 7.8 t 
any one of them may be considered as the basic theorem and the other 
two as corollaries of it. 

6. Prove Theorem 7.13 for the case in which a =5^ u and t = c. 

In Figure 7-32, lines o, b, c, and d are eoplanar and the measures of several 
angles are marked. In Exercises 7-9, justify the given assertion regarding 
the lines in this figure. 




Figure 7:12 



7, a is parallel to h. 

8. a is parallel to c. 
9* h is parallel to c. 

10. JSa t b,c are eoplanar lines such that a is parallel to b and b is parallel to 
c+ does it follow that a is parallel to e? Is this an instance of the reflexive 

property of parallelism for lines? Of the symmetric property? Of the 

transitive property? 

11* If a t b, are eoplanar lines such that a is parallel to b and b is parallel to 
c, docs it follow thai b is parallel to n? Is this an instance of the reflexive 
property of parallelism for lines? Of the symmetric property? Of the 
transitive property? 

12. If a, h. c are eoplanar lines such that a is parallel to b and b is parallel to 
C does it follow that c is parallel to c? Is tilts an instance of the reflexive 
property of parallelism for lines? Of the symmetric property? Of lite 
transitive property? 



306 



Paralii - 



Chapter 7 



In Figure 7-33, a, b t c, d are coplanar parallel lines, f is a transversal of ihcm, 
and the measures of se%'eral angles are marked. In Exercises 13-18, justify 
the given assertion regarding angle measure. 

13. x= 117 

14. y = 63 

15. z - 117 

16. u - 63 

17. o = 63 

18. w = 63 



af 



b + 



t + 




d + 



J 



Figure 7-3.1 

in F.xerciscs 19-21, use Figure 7-34 with E-A-D, m Z EAB = 1 IS, KD | BC. 



19. Find™ lABDHZm LABD = 3m Z.DBC. 

20. Find m /DBC and in ZBDA if 2m 
lABD = 3m I DBC 




22. Given a plane figure made up of two parallel lines, a transversal, and a 
segment, with angle measures as marked, find x\ y, and z. 




23. Given a plane figure made up of two pairs of parallel lines, with angle 
measures as marked, find x, y, z, u, v, and to. 




7.6 The Parallel Postulate 



307 



24. Given a plane figure with AB DE. with A. C, E all on the same side of 
BD, with C on the fc-side of AB, and with C on the A -side of ED, prove 



that 

m Z BCD = m / ABC + m Z COE 
(Hint- Copy the Figure and draw the line 
through C parallel to 12.) 



4 — *■ n 



25, In the figure for Exercise 24, what is the sum of the measures of Z ARC, 
LCBD, LBDC&ad /LCDE? 

26, In the figure for Exercise 24, what is the sum of the measures of Z CUD* 
ZBDC. and ZDCB? 

27, The figure helow shows some segments that are parallel and some that 
are not parallel. Try to write a good definition of parallel segments. 



Punt"!,-: 



Parallel Not pimiltet V Not paralM 




25. The figure below shows some quadrilaterals that are parallelograms and 
some that are not. Try to write a good definition of parallelogram. 




308 Parallelism : :: - ~ 



7.7 PARALLELISM FOR SEGMENTS; PARALLELOGRAMS 

In this section we introduce some definitions and theorems con- 
cerning an important type of quadrilateral called a parallelogram. Since 
the sides of a quadrilateral and, hence, of a parallelogram are segments., 
we need a formal definition that extends the concept of parallelism to 
segments. In Section 7,8 we extend the concept also to rays. 

Definition 7.1 If the lines which contain two segments are 
parallel, then the segments are said to lie parallel segments. 
and each is said to be parallel to the other. The segments in a 
set of segments are parallel if every two of them are parallel. 

The lines which contain parallel segments need not be distinct. In 
other words, a segment of a line is parallel to every segment of that 
line. As a special case of a special case, we note that every segment is 
parallel to itself. 

Let A, B, C t D be four points with A^B and C ^ D. {See Figure 

7-35.) Then KB is parallel to TJD if and only if AS is parallel to CD. We 
use the same symbol to denote parallelism for segments that we use for 
lines. Thus AB . CD means that AB is parallel to CD. 



C D 

-t ^i ► Figure 7-35 



Definition 7.8 A parallelogram is a quadrilateral each of 
whose sides is parallel to the side opposite it 

Consider a parallelogram ABCD as shown in Figure 7-36. Since 
ABCD is a quadrilateral, it follows that AB, BC, CD, DA are four dis- 
tinct lines- Since An and CD are distinct parallel lines, it follows that 
C and D He on the same side of AB* Similarly, D and A lie on the same 
side of BC, A and B lie on the same side of CD, and B and C lie on the 

same side of DA. Therefore each side of a parallelogram lies on a line 
which is the edge of a halfplane that contains all of the parallelogram 
D C 



Figure 7-36 



7.7 Pararffelitm for Segments; Parattetograrm 309 

except that side. Therefore every parallelogram is a convex quadri- 
lateral. (The word convex is used here in the sense of a convex polygon. 
See Definition 4, 1.6.) 

The next three theorems state some important properties of 
parallelograms. 

THEOREM 7.14 If a convex quadrilateral is a parallelogram, 
then its opposite sides are eongnient. 

Proof: Let parallelogram ABCD 
be^ given. (See Figure 7-37.) Then 
AB i| CD and BC \\ DA. We shall 
prove that AB = CD and that 
BC = DA. First draw the seg- 
ment A€, Our plan is to use con- 
gment triangles, 



Statement 



1, ZBACs IDCA 



% ACzsZA 

3. AACBsa LOAD 

4. AAiJCss &CDA 

5. ab ^ £E anc i bc ^ DA 




Reason 



1. Alternate interior angles de- 
termined by a transversal of 
two parallel lines are congruent, 

2. Why? 

3. Why? 

4. Why? 

5. Why? 



JHEOREM 7.15 If a convex quadrilateral is a parallelogram, then 
its opposite angles are congruent. 

Proof: Assigned as an exercise. 

THEOREM 7, 16 If a convex quadrilateral is a parallelogram, then 
its diagonals bisect each other. 

Proof: Assigned as an exercise. 

The next three theorems are useful in proving that certain quadri- 
laterals are parallelograms. 

THEOREM 7.17 If two sides of a convex quadrilateral are parallel 
and congruent, then the quadrilateral is a parallelogram. 



Proof: Assigned as an exercise. 



310 Parallelism 



Chapter 7 



TITEORFM 7J8 If the diagonals of a convex quadri lateral bisect 
each other, then the quadrilateral is a parallelogram. 

Proof: Assigned as an exercise. 

THEOREM 7.19 If each two opposite sides of a convex quadri- 
latend are congruent, then the quadrilateral is a parallelogram. 

Proof: Assigned as an exercise. 

Just as parallelograms are convex quadrilaterals with a special prop- 
erty, so are trapezoids convex quadrilaterals with a special prop- 
erty (only not quite as special as that for parallelograms). 



Definition 7.9 A trapezoid is a convex quadrilateral with 
least two parallel sides. 






Note that we do not say two and only two sides parallel. A trape- 
zoid may have only one pair of parallel sides or it may have two pairs 
of parallel sides. In other words, every parallelogram is a trapezoid, but 
not every trapezoid is a parallelogram. In some l*x>ks trapezoids are 
restricted to have only two parallel sides. In these instances the set of 
all trapezoids and the set of all parallelograms do not intersect. In this 
book the set of all parallelograms is a subset of the set of all trapezoids. 

Rhombuses, rectangles, and squares are all parallelograms with 
special properties. Their formal definitions come next. 



Definition 7.10 A rhombus is a parallelogram with two ad- 
jacent sides congruent 

Definition 7.11 A rectangle is a parallelogram with a right 
angle. 

Definition 7.12 A square is a rectangle with two adjacent 
sides congruent. 



THEOREM 7.20 A rhombus is an equilateral parallelogram. 
Proof Assigned as an exercise. 



7,7 ParalltHsm tor Segments; ParallefogTams 311 

THEOREM 7.21 A rectangle is a parallelogram with four congru- 
ent angles. 

Proof: Assigned as an exercise. 

THEOREM 7.22 A square is an equilateral rectangle. 
Proof; Assigned as an exercise, 

THEOREM 7,23 A square is an equiangular rhombus. 
Proof: Assigned as an exercise. 

THEOREM 7.24 The diagonals of a rhombus are perpendicular. 
Proof: Assigned as an exercise. 

In order to prove that a figure is a rhombus it is sufficient to prove 
that it is a parallelogram with two adjacent sides that are congruent 
(Definition 7.10). Tf we know that a figure is a rhombus, then we may 
conclude that all four of its sides are congruent (Theorem 7.20), If we 
want to show that a parallelogram is a rectangle, it is sufficient to show 
that it has one right angle, since it then necessarily has four right angles 
(Definition 7,11 and Theorem 7,21' lb show that a rect&ngfc is a 
square, it is sufficient to show that two of its adjacent sides are congru- 
ent, since then all four of its sides are congruent (Definition 7.12 and 
Theorem 7.22}. To show that a rhombus is a square, it is sufficient to 
show that it has a right angle since it is easy to show that an equilateral 
parallelogram with a right angle is an equilateral rectangle, that is, a 
sqiiare. 

In Chapter 3 we introduced the concept of distance. We restricted 
ourselves there to the idea of the distance between two points. We are 
ready now to extend the idea oi' distance to the distance between two 
parallel lines. Of course, we agree that the distance between a line and 
itself is zero. So let us consider the idea of the distance between two 
distinct parallel lines m and n as suggested in Figure 7-38. The length 
of a segment (sec the "dashed" segment in the figure) joining a point 
of one line to a point of the other line might be very long depending 
on bow the endpoiuLs are picked. 

« i v^ 1 1 



312 



Parallelism 



Chapter 7 



It seems natural to think of the distance between two parallel lines 
as the length of the shortest segment joining the two lines, and it seems 
that this segment should be perpendicular to both lines. If we pick any 
point on m, say M, then there is a line t in the plane of m and n that is 
perpendicular to m at M , (See Figure 7-39.) Also I intersects n in some 
point* say A 7 , and MN is perpendicular to both m and n. If t' is another 
transversal of m and n perpendicular to both of them and intersecting 
them in M' and -V, respectively, how do we know that M.N and M'N' 
are congruent? Our idea for the distance between two parallel lines 
will not be any good unless we can show that MN and M'S 7 are con- 
gruent. The next theorem shows that this is possible. 




Figure 7-39 

THEOREM 7.25 For every two distinct parallel lines there is a 
number that is the common length of all segments perpendicular to 
both of die given lines and with one end point on one of the given 
lines and one eiidpotnl on the other one. 

Proof: {See Figure 7-40.) I>et m and n be two distinct parallel lines. 

a b 




Figure 7-40 

Let A and B be two distinct points of m, and C and D two distinct 
points of n such that the segments AC and BD are perpendicular to 

both m and n. Then m is a transversal of AC and BD and is perpendicu- 
lar to both. It follows that KC and WD are parallel (Theorem 7.2); hence 
ABCD is a parallelogram (by definition) and AC = BD (Theorem 7.14), 

If we think of segment KC in Figure 7-40 as a fixed segment and BD 
m a variable segment (one that we can pick anywhere as long as B is 
on m, D is on n, and ¥D is perpendicular to m and to n), then the num- 
ber we are looking for, the one whose existence wc wanted to prove, is 
the number AC. The idea of Theorem 7,25 is sometimes expressed by 
saying that two parallel lines arc everywhere equidistant. 

We are now ready for the following definition. 



7.7 Parallelism for Segments; Parallelograms 313 



Definition 7.13 The distance between two distinct parallel 
lines is the length of a segment which is perpendicular to both 
lines and whose endpoints lie on these lines, one endpoint on 
one line and the other endpoint on the other line. The dis- 
tance between a line and itself is zero. 



EXERCISES 7.7 



1. Frove: If the convex quadrilateral ABCD is a parallelogram, then Z A fix 
Z C and I B as ID. (This is Theorem 7.15.) 

2. Prove: If the convex quadrilateral ABCD Is a parallelogram, then A~C 
and WD bisect each other. (This is Theorem 7.16.) 

3. Frove: II ABCD is a convex quadrilateral, if AB m CH5, and if AB || CD, 
then ABCD is a parallelogram. (This is Theorem 7.17.) 

4. From: If ABCD is a convex quadrilateral, and if AC and #D bisect each 
other, then ABCD is a parallelogram. (This is Theorem 7.18,) 

5. Frove: If ABCD is a convex quadrilateral if AB ==? CD and All as BC t 
then ABCD is a parallelogram. (This is Theorem 7.19.) 

6. If ABCD is a rhombus and if AB =s AD, then AB = BC = CD = DA 
and ABCD is an equilateral parallelogram. {This is Theorem 7.20.) 

7. Prove Theorem 7.21. 9. Prove Theorem 7.23. 

8. Prove Theorem 7.22. 10. Prove Theorem 7.24, 

11. Why are opposite sides of a parallelogram parallel? 

12. Why are opposite sides of a parallelogram congruent? 

13. The figure below shows some rays that are parallel and some that are 
not parallel, Try to write a good definition of parallel rays. 

A B c • > 4 • • » 



AB and BC are parotid Parallel rays Not parallel rays Not parallel nyt 

14. The figure below shows some antiparallel rays and some rays that are 
not antiparallel. Try to write a good definition of antiparallel rays. 

A B CD 
— • ► 4 — • — • • m 



BA and EC 

are antiparalle] 



* 



?Cot antiparaU*! 



BA and CD 

are antipartllel 



B 



ACnndAB 
are not aariparalM 



4- 



Antiparallr-l my* 




Not antiparaCe] 



314 



Parallelism 



Chapter 7 



7,8 PARALLELISM FOR RAYS 

In this section wo extend the concept of parallelism to rays. Since 
segments are parallel if the lines containing them arc parallel, we might 
consider rays parallel if the lines containing them are parallel. But it is 

useful to Distinguish between the case suggested by rays AB and CD 

and the case suggested by rays EF and GH in Figure 7-41. In both 
cases the rays lie on parallel lines. In one case, however, they point in 
the same direction, whereas in the other case they point in opposite 
directions. (We use the word "direction" here in an intuitive sense; it is 
not a part of our formal development*) 





Figure 7-41 

Definition 7A4 (Sec Figure 7-42.) Two noneollincar rays 

AB and CD are parallel if AB and CD are parallel lines and 

if B and D lie on the same side of AC. Two coDmear rays are 
parallel if one of them is a subset of the other. 



Definition 7,15 (See Figure 7-42.) Two noneollincar rays 

EF and GH arc anliparallel if EF and GH are parallel lines 

and if F and H lie on opposite sides of EG, Two collinear rays 
are an ti parallel if neither is a subset of the other. 





AB unci CD urv porulkl. 
Figure 7-42 



IJ*n±JK 
JIutAIKm 



iirti pontile]. 



EF mul GH are antiparalkL 



The proofs of tile following three theorems are assigned as ex- 
ercises. 

THEOREM 7.26 If Z A BC and I DEF are coplanar angles with 
BA and eB parallel and b8&xuI EF* parallel, then Z ABC^ Z DEF. 



7.8 Paraltellsm for Rays 315 

THEOREM 7.27 If Z ABC and t DEE are coplanar angles with 

— * — * ■ — ^ — * 

BA and ED parallel and with BC and Eb antiparallel, then / ABC 

and LDT.E are supplementary. 

THEOREM 7.2S If / ABC and Z OFF am coplanar angles with 

BA and ED antiparallel and with BC and EF antiparallel then 
LABC^ A DEE 



EXERCISES 7.S 

Dra%v a line and on it mark five points A, B, C, D, £ in the order named. In 
Exercises 1-20, slate whether the given .statement is true or fake. 



2. AB = 52 

3. 31 =s M 

4. AB = M 
5, AB || BA 

6. AB || AC 

7, IS || CD 

&.2Sc3ft 

9. AB c BC 
10, a2 C AC 



ilaScbS 

12. 5? C AB 

13, Ai! and BA arc parallel. 
14* An and BA are anti parallel. 
13, AB and FD are parallel. 

16. HC and BA are antiparallel. 

17. b? n bI = {B} 

18. Bf? f| DA = BD 

19. BC? P DA = Wi 

20. d£ 1CD= 



A t B T C, . . . , K, L are distinct coplanar points, as in Figure 7-43, so that 

t j -^ ) < | ^ j 

DF | j/i and D/f | Fi. In Exercises 21-30, state whether the given rays are 
parallel or antiparallel, 

21. 15i and 77? 

22. DE and 77? 

23. £// and BE 

24. I1A and Li 

25. ^andEF 

26. CF and FC 

27. CF andlf 

28. Cpmidjf 

29. Gland 77f 

30. GfmidjH 




Hfure7-43 



316 Parallelism Chapter 7 

■ In Exercises 31-36, copy and complete the statements. 

31. The intcrseetitm of two parallel collinear rays is [?]. 

32. The intersection of two parallel noncollinear rays is [7], 

33. The intersection of two antiparallel collinear rays is (?] or [Tj or (?]. 

34. The union of two parallel collinear rays is (t). 

35. The union of two anti parallel collinear rays is a line with the interior 
points of a segment deleted or it is an entire fT|. 

36. If ABVD is a parallelogram, then An and CD are [7] rays. 

37. Prove Theorem 7.26 for the case in which BA and ED are collinear and 
parallel and BC and JSFftrc noncollinear and parallel. (See the figure be- 
low.) The case in winch BA and ED are noncollinear and parallel and 
BC and EF are collinear and parallel can be proved in a similar way. 




38. Prove Theorem 7.26 for the case in which BA and ED are noncollinear 

and parallel and BC and £F arc noncollinear and parallel as suggested 

in the figure. (Hint-. Think of the figure formed by the lines Afi, BC, DE. 

and EF. Do you see that this figure contains a parallelogram? Then use 
Theorem 7.15.) 



39. In Exercises 37 and 38, you were asked to prove Theorem 7.26 except 

for the case in which BA and ED are collinear and parallel and BC and 

EFare collinear and parallel. Draw a figure for this special case. Note 
that in this special case the theorem amounts to saying that equal angles 
are congruent angles. 

40. Draw an appropriate figure and prove Theorem 7.27 for the case in 

which BA and ED are collinear and parallel and BC and EF are collinear 
and antiparallel. 

4L Draw an appropriate figure and prove Theorem "7.27 for the case in 

which BA and ED are noncollinear and parallel and BC and EF are 
collinear and antiparallel, 



7.8 Parallelism for Rays 317 

42. Draw an appropriate figure and prove Theorem 7,27 for the case in 

which BA and ED are collinear and par&tlel and BC and EF are non- 
coUinear and antiparallel. 

43. Draw an appropriate figure and prove Theorem 7.27 for the ease ia 

which BA and ED are noncollmear and parallel and BC and £F are noo- 

coDinear and an tj parallel, 

44. Prove Theorem 7.28. Do it by eases. In which special case does the as- 
sertion of the theorem amount to an assertion that vertical angles arc 
congruent? 

45. ABCD is a convex quadrilateral. Prove: If 

m£A +m/.B + mlC + m£D = 360, 

ifmZA a mZ C, andif mZB = roZ D t thenABCDis a parallelogram. 

46. Prove: If Z A and Z B are consecutive angles of a parallel ©gram, then 
they are supplementary. (How many pairs of consecutive angles does a 

parallelogram have?) 

In Kxercises 47-52, write a sentence justifying the truth of tho given 
statement. 

47. If two alternate interior angles determ ined by a transversal of two given 
copianar lines are not congruent, then the given lines are not parallel. 

48. If two corresponding angles detennined by a transversal of two given 
copianar lines are not congruent, then the given lines are not parallel. 

49. If two consecutive interior angles determined by a transversal of two 
given copianar lines are not supplementary, then the given lines are not 
parallel. 

50. If the diagonals of a quadrilateral do not bisect each other, then the 
quadrilateral is not a parallelogram. 

51. If two opposite sides of a quadrilateral are not parallel, then the quadri- 
lateral is not a parallelogram. 

52. If ABCD Is a parallelogram, then AB and CD are everywhere 

equidistant. 

53. Given two alternate interior angles detennined by a transversal of two 
parallel lines, prove that the angle bisectors of these angles are 
antiparallel ■ 

54. 'Hie figure shows five copianar 
lines and some (perhaps all} 
of their points of intersection. 
If m^L ABE + mLBEA + mLEAB 
= 180, I BAE at LEAC, Z ABE 
b Z DBE t and m Z BEA = 90, prove 

that BD is parallel to AC. 




318 Parallelism Chapter 7 

7.9 SOME THEOREMS ON TRIANGLES AND QUADRILATERALS 

One of the most important consequences of die Parallel Postulate 
is the following theorem. 

THEOREM 7«$9 The sum of the measures of tlie angles of a tri- 
angle is 180. 

Proof* (See Figure 7-44,) Let A ABC be given. Let m be the unique 

line through C and parallel to AB. Let D and E be points on m such that 

D-C-E and such that B and D are on opposite sides of AC as in the 
figure. Then 

mLA + mLB + mlACB 

= mlACD + m L BCE + mZ ACB = 1 SO. 

(Which theorems justify these equations?) 




It may be of interest to mention here that in the non-Euclidean ge- 
ometry of Lobachevsky and Bolyai the sum of the measures of the an- 
gles of a triangle is less than 180, The sum is not the same number for 
all triangles in that geometry. The smaller the triangle is the closer die 
sum is to 180. 

THEOREM 7,30 The measure of an exterior angle of a triangle is 
equal to the sum of the measures of its nonacljaeent interior angles. 

Proof: Assigned as an exercise, 

THEOREM 7,31 Let a one-to-one correspondence between die 
vertices of two triangles be given. If two angles of one triangle are 
congruent, respectively, to the corresponding angle* of the other 
triangle, then the third angles of the two triangles are also 
congruent. 



7.9 Theorems oti Triangles and Quadrilaterals 319 



Assigned as an exercise- (Note Figure 7-45 which suggests that 
the correspondence need not be a congruence. One of the triangles 
may be larger than the other.) 





Figure 7-45 

THEOREM 7.32 {The S.AJi. TJieorem) Let a one-to-one cor- 
respondence between the vertices of two triangles be given. If two 
angles and a side opposite one of them in one triangle are congru- 
ent, respectively, to the corresponding parts of the second triangle, 
then the correspondence is a congruence. 



Proof: 



Assigned as an exercise. 



THEOREM 7.33 The sum of die measures of the angles of a con- 
vex quadrilateral is 360. 

Proof: Let ABCD be a convex quadrilateral as in Figure 7-46. Draw 
diagonal BD. Since ABCD is a convex quadrilateral, D is in the interior 
of Z ABC and B is in the interior of Z CDA. Then, using the notation 
of the figure, we have 

mLA + mZABC + mLC + mLCDA 

= mZA -r {mil + m/2) + m£C+(mZ4 + mZ3) 
= (m IA + m Z 1 + m Z3) + m Z2 + (m LC + m Z4) 
= 180 + 180 = 360, 




THEOREM 7.34 The acute 
complementary. 



es of a right 



e are 



Proof: Assigned as an exercise. 



320 



Parallelism 



Chapter 7 



THEOREM 7.-35 (The Hypotenuse-Leg Theorem) Let there be 
a one-to-one correspondence between the vertices of two right 
triangles in which the vertices of the rigfrt angles correspond. If the 
hypotenuse and a leg of one triangle are congruent to the corre- 
sponding parts of the other triangle, then the correspondence is a 
congruence. 

Proof: (See Figure 7-47.) Let A ABC with mZA = 90 and ADEF 
with m£_D = 90 be given. Let it be given that BC^EF and 
AB = DE. We shall prove that ABC * — > DEFis a congruence. You 

will be asked in Exercise 19 below to supply the missing reasons, 





tatement 



1. Let G be a point on opp DF 

such that DG s AC, 
'2, mLEDG = 90 



3. ED = BA and ml. A 

4. ADEG =s AABC 

5. EG^BC 

6. BC = EF 

7. EG a EF 

8. ZEGDa ZKFD 

9. AOEG a ADEF 
10. AABCss ADEF 



- <.K) 



Reason 

1. The Segment Construction 
Theorem 

2. Z £Z>G is a supplement of a 
right angle 

3. Given 

4. S.A.S. 

5. a 

6. Given 

7.Q] 

8. E 

9. S.A.A. 

0. The equivalence properties 
of congruence for triangles 



EXERCISES 7.9 



In Exercises 1-6, the measures of two angles of a triangle are given. Find 
the measure of the third angle of that triangle, 

1. 93 and 80, 

% a and b. (Express the answer in terms of a and h) 

3. 25 and x. (Express the answer in terms of x.) 

4, 90 + k and 90 — 2k, (Express the answer in terms of Jt.) 



7.9 Ttworems on Triangles and Quadrilaterals 321 

5. 45 + x and 45 ■+- *. (Express the answer in terms of x.) 

6. 60 + h and 60 -+- t-. (Express the answer in terms of u and V.) 

In Exercises 7-ll t the measures of the three angles of a triangle are ex- 
pressed in terms of a number %. In each case, find the value of .v and cheek 
your answer by adding the anjiie measures. 



7. x, x — 5, x -s- 5 

S. x, 2x, 3x 

9. x + 1, x + 2, x + 3 



10. x + 50, 2x + 30. 3x + 80 

11, 3x + o»2r - 3, ox — 8 



12. The figure shows a triangle A ABC, a point D between A and tf, the line 

CD, and the lines parallel to CD through A and B. What is the sum of 
the measures of angles 1 , 2, 5, and 6? How does the sum of the measures 
of Z 1 and Z 6 compare with the sum of the measures of £ 3 and Z4? 
How docs the sum of the measures of / 3 and Z 4 compare with the 
measure of £ACB? Use these ideas to write an alternate proof of The- 
orem 7.29, 




13, The figure shows a parallelogram ABCD and one of its diagonals AC, 

D C 




What is the sum of the measures of angles LDAB and Z B? Why? 
What is the sum of the measures of angles LDCB and I D? 
What is the suni of the measures of the angles of a parallelogram? 
Why is AB ss CV, m m Z3A~ t and 55 SM A~C? 
Why is A ABC s ACDA? 

Use these ideas to write an alternate proof of Theorem 7.29. 
Prove Theorem 7,30, 
Prove Theorem 7.31. 
16. Prove Theorem 7.32. (Start with two triangles, &ABC and AA'JS'C, 
and a correspondence ABC <s — > A'B'C) 



It. 
15. 



322 Parallelism Chapter 7 

17. Justify the equations at the end of the proof of Theorem 7.33. 

18. Prove Theorem 7.34. 

19. Copy and supply reasons for steps 5, 7, 8 in the proof of Theorem 7.-35. 

20. Let A ABC; and uurntan a, b. c be given. Suppose that m£A = ka, 
m£B = kh, m L C = kc for some number A% the same number k in all 
three cases, If a = 1,6 = 2,0 = 3, find mZA, mlB, mjLC. Check 
by addition. 

21. Same as Exercise 20, except that a = 2, h = 3, c ss 5. 

22. Same as Exercise 20, except that a = 100, h = 250, c = 300. 

23. Same as Exercise 20, except that a s 180, h = 180, a = 180. 

24. Same as Exercise 20, except that fits 1, h = 1, c = 100. 

25. Same as Exercise 20, except that ft = 100, b = 1 , c = 100. 

■ In Exercises 26-30. there is a labeled figure in which x denotes the measure 
of an angle and there is an assertion about .t. Justify the assertion. In the fig- 
ure for Exercise 30, y and z also denote angle measures. 



20, i = 50 



x = 90 





27. ;v = 150 



29. x = 115 




30. x = 27 



7,9 Thwrtms wi Triangles and Quadrilaterals 323 

In Exercises 31-35, Ihcre is a labeled figure in which x denotes the length of 
a segment. At the left, there is an assertion about x. Justify the assertion. If 
ij appears in a figure, it also denotes a length. 



31. x = 12 



32. X = if 



33. x = 10 



34, X= 11 



35. x = y 




36. If the measure of the vertex angle of an isosceles triangle is 12.76, find 
the measure of each l>ase angle. 

37. If the measure of each base angle of an isosceles triangle is 83, Dnd the 
measure of an exterior angle at the vertex of the triangle. 

38. Find the sum of the measures of the exterior angles (one ul each vertex) 
of a triangle whose angles have measures as follows: 27,32* 59.39, and 
93.2& 

39. Given a parallelogram A BCD with AC — BD, find the measures of the 
four angles of the parallelogram. 

40. Can a convex quadrilateral have three obtuse angles? Explain your 
answer. 



324 Parallelism Chaptef ? 

41. Given a convex quadrilateral ABCD with mZA = 90, find mLB + 
tnZC + mZD. Can LB and Z C both be obtuse angles? 

42. Let AiJCD be aeon vex quadrilateral with m Z A = 90andmZB = 90. 
Can Z C and Z D both be obtuse angles? 

43. L«t AS and CD intersect at E, an interior point of each segment. If 
AD = BC and AD || W, t prove that DC and AH bisect each other. 

44. The measures of the angles of a triangle are 36, x t and y. We conclude 
that x is a number less than [7], tliat y is a number less than {T\, and that 
x + If = 13 ( In each case find the smallest number that will make the 
statement true.) 

45. The measures af the angles of a convex quadrilateral are a, b> c, d. We 
conclude that each of these four numbers is less than [T|. (Find the 
smallest number that will make the statement true.) 

46. The measures of the angles of a convex quadrilateral are 20, x, x, and y. 
We conclude that x is a number between [T| and [T|. that y is a number 
between £TJ and jT], (Make the smaller number as large as possible and 
the larger number as small as possible in each case.) 

47. Given the plane figure in which AC is the bisector of Z BAD, CB X AB S 
and CD _L AD, prove that BC = DC. 




48. Prove: If each angle of a convex quadrilateral is congruent to the angle 
opposite to it, then the quadrilateral is a parallelogram. [Hint: Given 
quadrilateral A BCD with LA^ LC % LBst Z£) } find»iZA + m/ H 
and deduee that two sides are parallel. Find »«ZA + mLD and de- 
duce that the other two sides are parallel.) 

49. Given A ABC and A A DC with JD intersecting EC at M t an interior 
point of AD and EC as indicated in the figure, find the angle measures 
x, y, 3 u, t\ 




R«view Exercises 325 

CHAPTER SUMMARY 

The central theme of this chapter is parallel lines, and the Iiigh point 
Of I he chapter is ihe PARALLEL POSTULATE. Before introducing Lhe Par- 
allel Postulate we proved several theorems on conditions which imply that 
lines are parallel These arc theorems in EUCLIDEAN GEOMETRY as 
wdl n in the NON-EUCLIDEAN GEOMETRY of Lobachcvsky and 
Bcdyai. These theorems are concerned with lines and TRANSVERSALS and 
angles associated with them. The Parallel Postulate gives us what we need to 
prove the converses of these theorems. 

PARALLELISM for lines in a given plane is an equivalence relation. 
that is, it is reflexive, symmetric, and transitive. The same is true for paral- 
lelism of segments in a given plane and for rays in a given plane. 

With parallel lines it is natural to associate PARALLELOGRAMS. 
This chapter includes several theorems regarding properties of parallelo- 
grams and several others regarding conditions tinder which quadrilaterals 
are parallelograms. 

An important consequence of the Parallel Postulate is the theorem on 
the sum of the measures of the angles of a triangle. An easy corollary of this 
theorem is the theorem on the sum of the measures of the angles of a convex 
qnadrilateraL Another important consequence of the Parallel Postulate is 
the result that parallel lines are everywhere equidistant 

Along with the idea of PARALLEL RAYS we introduced the idea of 
A NT! PARALLEL RAYS, There were several theorems regarding the re- 
lationship of two angles with parallel or anti parallel sides. 

Early in the book we introduced the S.A.S., A.S.A., and $.$*& Postu- 
lates. In this chapter there were two results which extend these congruence 
ideas, namely the S.AA. THEOREM and the HYPOTENUSE-LEG 
THEOREM for right triangles. 



REVIEW EXERCISES 

Copy and complete the definitious in Exercises 1-10. 

I, line m is parallel to line n if and only if [TJ, 

SL Two lines are skew lines if and only if (TJ- 

3, Two segments are parallel if and only if \T\. 

A. Two collinear rays arc parallel if and only if jT|. 

5. Two noncollincar rays are parallel If and only if [fj, 

& Two collinear rays arc antiparallel if and only if JT], 

7. Two noncoUinear rays are antiparallel if and only if '|T]. 

8. If distinct lines m, n t t are coplanar, then Ms a tramversal of m and n if 
and only if [TJ, 



326 Parallelism Chapter 7 

9. A parallelogram is a convex quadrilateral whose [?j. 
10. The distance between two distinct parallel lines is [?]. 

■ Exercises 1 1-20 pertain to the geometrical figure suggested by Figure 7-48. 

It is given that lines A& % CD, and EF arc parallel and noncoplaaar. Also, 

AB = CD = EF. 

In each exercise tell whether llic given statement is true or false, and explain 
why it is true Off why it is false. 




Figure 7-45 £ 

11. There is more than one plane contacting A and B. 

12. There is more than one plane containing A, B> and C. 

13. There is exactly one plane containing A, B> E, and R 
14 There is exactly one plane containing A, £, C, and F. 

15. Quadrilateral ABDC is a parallelogram. 

16. Quadrilateral ABFE is a parallelogram. 

17. Quadrilateral CDFE is a parallelogram. 

18. AACE St ABDF 

19. A ABC =s ADCfl 

20. m / £4 C+ in lACE+m£ CEA + m L FBD +ml. BDF+ m Z DVB 
= wZ CAB + m Z A£D+ mZ BDC+ m Z DCA 

Exercises 21-30 pertain to the geometrical figure suggested by Figure 7-49. 
Line ( is a transversal of lines m and n, The measures of eight angles formed 
by m, n, and f. have been marked in the figure. In each exercise, tell whether 
tile statement is true or false. If it is true* state a theorem that justifies your 
answer. 



Figure 7-49 




Rftvtaw Exercises 327 

21, If * = 125, then in [\ n, 26. If m ]n, then w = 125. 

22, If u = 65, then in || n. 27. If n = 60, then m is not parallel to n. 

23, If w = 125, then m | n, 28. If .v = MO, then m is not parallel to n. 

24, If m '| n, then 3 = 125. 29. If y =£ u, then m is not parallel to n. 

25, If m | n» then u = 65. 30. x = 126 



31. In your own words write a sentence that tells of a significant difference 
between the geometry of Euclid and the geometry of Lobachevsky and 
Bolyai. 

32. Explain what we mean when we say that the equals relation is reflexive, 
synunetric, and transitive. 

33. Explain what we mean when we say that the relation of parallelism for 
lines is reflexive, symmetric, and transitive. 

34. Is the relation of parallelism for rays an equivalence relation? 
;15. h the relation oi antipardllelism For rays on equiva'.ru.-..- rln.n m.- 

In Exercises 36-42, you are asked to prove a statement. You may use any- 
thing that we have had in our formal geometry structure up to this point 
(except the theorem you are asked to prove) in writing your proof. 

36. Prove that opposite sides of a parallelogram are congruent. 

37. Prove that if two opposite sides of a convex quadrilateral are congruent 
and parallel, then the quadrilateral is a parallelogram. 

38. Prove that if opposite sides of a convex quadrilateral are congruent, then 

the quadrilateral is u parallelogram. 

39. Prove that opposite angles of a paralletogram are congruent 

40. Prove that two consecutive angles of a parallelogram are supplemen- 
tary. 

41. Prove that the two acute angles of a right triangle arc complementary 

angles. 

42. State and prove the theorem regarding the angle measure sum of a 
triangle. 

43. In the geometry of lobachevsky and Bolyai is it tnie that two alternate 
interior angles formed by two parallel lines and a transversal are 

congruent? 

44. In the geometry of Euclid is it true that two alternate interior angles 
formed by two parallel lines and a transversal are congruent? 

45. In the geometry of Lohachevsky and Bolyai is it true that it' two alter- 
nate interior angles formed by two distinct lines and a transversal of 
them are congruent, then the lines are parallel? 

46. In the geometry of Euclid is it true that if two alternate interior angles 
formed by two distinct coplanar lines and a transversal of them are con- 
gruent, then the lines are parallel? 





Chapter 



Fritz Henlc/Fhoto Researclien 



Perpendicularity 
and Parallelism 
in Space 



8.1 INTRODUCTION 

This chapter is concerned mostly with figures that do not lie in one 
plane. We use the word figure in two ways: as a set of points (geo- 
metrical figure) and as a picture or drawing that represents a geo- 
metrical figure. The figures in this book and the figures that you draw 
as you study geometry are important. In communicating information 
regarding geometrical figures one drawing may be worth 473 words. 

When drawing a figure, the perspective view is the t>cst one from 
the standpoint of communicating how the figure looks, and that is the 
view used in the illustrations in this book. However, the oblique view 
is the easiest to draw; therefore instructions for drawing figures in 
oblique views are given on the following two pages. 

A plane may be suggested by a parallelogram as in the figure 
which illustrates two intersecting planes. Note that the plane sug- 
gested by a parallelogram includes more than a parallelogram and its 
interior. It includes all of the points in the plane of lire parallelogram, 
the points in the exterior of the parallelogram as well as those on the 
parallelogram and in its interior. 



PROCEDURES FOR 

DRAWING GEOMETRIC 

FORMS IN 

OBlfQUE. VIEWS 






Draw front plane, edge 
or edges of form, in 
norma* elevation view. 




Piojeti from Hie 
comers, al a 45* 
angle, facetting 

Lid'd k h'..-,, wllos= 

unit of measurement 
sl Diid m H of those 
in Iht' elevation view. 



The bteck dashes 
ore guide lines to 
b* crated later. 





z 






Draw vertical gu.de 
lm« for the center 

of tne pyramid. 



/ / 



y 



y* 76 



330 







A 



7 / 




£JZ7 



/ 




ines that aro Inside 
or bor.irtd an obfect 
are by convention 
drawn 35 dashes. 





331 



332 Parpandlcularity and Parallelism Jn Space 



Chapter 8 



A horizontal plane is oriented like the top of the cube. A vertical 
plane is oriented like the right face of the cube. Note Lhat horizontal 
and vertical are not defined formally in our geometry. They arc 
descriptive terms that we use to describe figures. 

The main topics in this chapter are perpendicularity and parallel- 
ism for lines and planes, in the preceding chapters of this book we 
have given detailed proofs for most of the theorems that were stated. 
Proofs not included in the text were assigned as exercises. In this 
chapter, however, we sometimes state theorems without giving proofs 
and without assigning these proofs in the exercises. In such cases we 
may include an outline of the main steps or a short statement of the 
strategy or main idea of a possible proof. In all cases it is possible to 
write detailed proofs based on the postulates of our formal geometry. 
Our abbreviated presentation is appropriate for a first course in formal 
geometry. 

8.2 A PERPENDICULARITY DEFINITION 



Definition 8.1 A line and a plane are perpendicular if the 
line intersects the plane and is perpendicular to ever)' line in 
the plane through the point of intersection. 



If a line / and a plane a are perpendicular, we say that I is perpen- 
dicular to a and that a is perpendicular to /. 

Notation, / 1 a, or a 1 /, means that a and / arc perpendicular; 
I J_ a at P, or a 1 2 at P, means that t X a and that P is their point of 
intersection. 

Figure 8-1 represents a line / and a plane a that are perpendicular 
at P. The figure shows four of the lines through P and in a. Actually, 
there are infinitely many such lines and they are all perpendicular to 
I at P. 




riffureS-i 



9.2 A Perpendicularity Definition 333 



EXERCISES 3.2 

■ In Exercises 1-9, draw the figure that is described by the following 
statements. 

1, Two distinct horizontal planes, 

2. Two distinct vertical planes. 

& A horizontal plane and a vertical line, 

4, A vertical plane and a horizontal tine. 

5. A vertical plane and a line perpendicular to it 

6* Two distinct parallel planes and a line that intersects both of them, 

7. Two distinct parallel lines and a plane that is parallel to both of the lines. 

8. Two distinct intersecting planes, one vertical plane and one horizontal 
plane. 

9. Two congruent triangles tying in distinct parallel planes and the seg- 
ments connecting corresponding vertices, 

10, See the figure. If i _L m, and m lies in a. docs it follow that / J_ a? 
Rxplain 




1L In the figure, A-B-A\ AB = A% aX' I a at B t B $6 C, and C £ a. 
Prove that 

AC = A'C 



A 



\ 



! / 



A' 



334 Perpendicularity and Parallelism in $p*ce Chapter 8 

12, In the following figure, aA' j. a at P, A'-P-A, A? = A'P, and B-D-C. 
On the basis of Exercise 1 i f what can you say about the lengths AB and 
A% AC and A'C, AD and A'D? Now prove that AABD = AA'BD, 
AACD s= AA'CD, A ABC s AA'JSC. 




13. In the figure in Exercise 12, AT ± PB, AP ± PC; A'-P-A; B-D-C; a 
is the plane containing the noncollinear points B, C, P; AB — A'B; 
AC = A'C. State the postulate which implies that D is in a, Prove that 
AD := A'D. 

14. {An informal geometry exercise*) Let I and m be two distinct lines in 
the plane of a table top and intersecting at a point R Hold a yardstick 
so that it appears to be perpendicular to I and to in at P. Use another 
yardstick to represent a third line n in the plane of the table top and 

passing through P. Docs the first yardstick appear to be perpendicular 
to lino n? Try n in various positions and see if you can find a position 
such that the yardstick and line n are no longer perpendicular. Does it 
appear that the yardstick is perpendicular to lino n in all cases? Does it 
appear that the yardstick is perpendicular to the table? 



8.3 A BASIC PERPENDICULARITY THEOREM 

Theorem 8.1 is introduced to help us prove a basic perpendicularity 
theorem, Theorem 8.2. Theorem 8.2 is suggested by our experiences 
with perpendicular lines and planes. See Exercise 14 of Exercises 8.2, 

THEOREM 8. / If A, A ', B, C, D are distinct points with B and C 

each equidistant from A and A' and with D on BC, then D is equidis- 
tant from A and A f . 



Given distinct points A, A\ B, Q A with AB = A'B, AC = 

A'C, and D on BC as suggested in Figure 8.2, we want to prove that 
AD s A'D. Then D-B-Q or B-D-C 7 or B-C-D. The proof may be 
completed by showing that A ABC ^ AA'BC* AABD =* AA'BD* 
AACD a AA'CD. and hence that AD = A'D* 



8,3 A Basic Perpendicularity Theorem 335 




qp»u 

THEOREM 8.2 If a line is perpendicular to each of two distinct 
intersecting lines at their point of intersection, then it is perpen- 
dicular to the plane that contains them. 

Proof: Let lines l r m T n be given with m, n in plane a and such that 
I _L m at A y I 1 natA, as suggested in Figure 8-3. Let p be any line 
distinct from TO and n that ties in a and passes through A. We want to 
prove that I 1 p. Let q he a line in a that i s not parallel to m, or to n, or 
to p. and that docs not pass through A. Then q intersects to, p* rt in 
three distinct points; call them M, P, N, respectively. 

Let Q awl (r be two points of I such that Q-A-Q' and QA = Q'A, 
The proof may be completed by showing that M is equidistant from 
Q and Q\ S is equidistant from Q and Q', P is equidistant from {) 
and Q'j 

AQPA a AQ^PA, LQAP^ LQ'AP, and I X p. 

Since p is an arbitrary line oilier than to and n in the plane a and pass- 
ing through the point A, we have I X a. 




Vigun W 



336 Perpendicularity and Parallel ism in Space 



Chapter 8 



EXERCISES 8.3 

1. Let the distinct points A, B, C, A E be given as suggested in the figure. 
A, B r C. D are coplanar points with no three of them eollinear. Also 
ED I W and ED 1 DC, Prove that ED 1 M. 



.?* 




2. Let the distinct points A, H> C t D be given as suggested in the figure 

A, /*, C are noncollinear points in a plane «, S3 I a, and JJ = #C, 
Prove that IDAC a ZDCA, 




3. Let the distinct points A, S t C, £> be given as suggested in the figure, 
ATS 1 J57), and AB _L HC. A, B, C are noncollinear points is a, and I) 



4—* 



is a point not in a. Prove that AD is not perpendicular to a. 




8.3 A Basic Perpendicularity Theorem 337 

4, For the situation in Exercise 3, prove that AD is not perpendicular to 

tlie plane determined hy B, C D. Is AjB perpendicular to the plane 
BCD? 

5, Let four noncoplan&r points A> B, C, D be given such that 

AB = AC = AD, 

A3 i plane ACD, AC 1 plane BAD, A& X plane BAG. Draw an ap- 
propriate figure and prove that A BCD is an equilateral triangle. 

6, A rectangular solid has the property that if two of its edges intersect, 
then they intersect at right angles. The following figure shows a rcctan- 

gular solid with the eight vertices labeled. Prove that EA is perpendicu- 
lar to the plane that contains B t C, D. 




7, Assume the same situation as in Exercise 6. Since EA is perpendicular 
to the plane that contains the quadrilateral ABCD t wc say that edge EA 
is perpendicular to face A BCD. Name six other combinations of an edge 
and a face that are perpendicular. (There are 24 such combinations 
altogether.) 

8, In the proof of Theorem 8.1, suppose that D-B-C. Which Congruence 
Postulate would you use to show that AABCs AA'BC? That 
AABD Sf AA'BD? 

9, Explain how the proof of Theorem 8.1 for the case in which B-D-C 
differs from the proof for the case in which D-B-C, Draw an appropri- 
ate figure. 

10. Draw a figure like Figure 8-3, except much larger. Draw the segments 
connecting Q and Q with M, P, N. In completing the proof of Theorem 
8.2, which theorem, postulate, or definition plays a key role in proving 
that 

(ft) M is equidistant from Q and Q' and that N is equidistant from Q 
and Q'? 

(b) F is equidistant from Q and p? 

(c) &QPA s AQ'PA? 

(d) LQAF=z Zp'AP? 

(e) / 1 p? 



338 Perpendicularity and ParaKtali&m in Space 



ChaplerS 



8,4 OTHER PERPENDICULARITY THEOREMS 

THEORFM 8,3 If a line and a plane are perpendicular, then the 
plane contains every line perpendicular to the given line at the 
point of intersection of the given line and the given plane. 

Proof: {See Figure 8-4.) Let / I a at P. Let t x be any line such that 
/, ^/atP. 




Figured 



We want to prove that ^ lies in «, Let be the plane that 
contains / and ly (See Figure 8-3.) Then /? ^ a since contains / and 
g does not. The intersection of (i and a is a line; call it / 2 . Then 1% _L J. 
Why? Also h ± I Why? Then h = k* Why? Therefore /, lies in a. 
Since ^ is any arbitrary line perpendicular to / at F, die proof is 
complete. 




figure. S-5 



8,4 Other Perpendicularity Theorems 339 

THEOREM 8.4 Given a line and a point, there is a unique plane 

perpendicular to the line and containing the point 

Proof: Lei £ be a given line and P a given point. We divide the proof 
into four parts. 

1. If P £ /, we must prove that there is 
at least one plane perpendicular to / 
at P. 

2. If P £ J, we must prove that there is 
at most one plane perpendicular to I 
at R 

3. If P £ /, we must prove that there is 
at least one plane perpendicular to I 
and passing through P. 

4. If P £ I, wc must prove that there is 

at most one plane perpendicular to I 
and passing through P. 

Parts 1 and 3 are existence proofs; parts 2 and 4 are uniqueness 



Proof of 1; If P is a point on a line 2 as in Figure 8-6, let rt be a plane 
containing I and let in be the line in a that is perpendicular to I at P. 




Figure Wl 



I^t Q be a point not in a and let fi be the plane that contains Q and L 
Let n be the line in /J that is perpendicular to / at P. Let v ta the plane 
that contains m and n. Then I A. y at E Why? This proves that there 
is at least one plane perpendicular to 1 at P. 



340 Ptrpandteularity and Parallelism In Spac* 



Chapters 



Proof of 2: If point P lies on a line I as in Figure 8-7, we shall prove 
that there is at most one plane perpendicular to / at P. Suppose, eon- 




Figure S-7 

trary to what we assert, that there are two distinct planes a and /? each 
perpendicular to I at P. Then a and /? intersect. Why? Their intersection 
is a line. Why? Call this line m. Let f and s be the lines such that r is in 
a and rimat F, and * is In /? and s 1 m at P, Let y be the plane that 
contains r and s. Then m J_ y at P (Why?), m X I at f (Why?), and I 
lies in y. Then % r, s are three distinct eoplanar lines with r _ I and 
s J. I, But this is impossible, There cannot be two distinct planes each 
perpendicular to I at P. Therefore there is at most one plane perpen- 
dicular to I at P, 

Proof of 3: Given a line / and a point P not on I as in Figure 8-8, let 
Q be the foot of the perpendicular from P to I Let a be the unique 
plane such that a ^_ I at Q. How do we know that there is one and only 
one such plane a? Then P lies in a. Why? Therefore there is at least 
one plane containing P and perpendicular to I . 




Figure B-9 



8.4 Other Perpendicularity Theorems 341 

Proof of 4: Given a line I and a point P not on it, as in Figures 8-9 
and 8-10, we shall prove that there is at most one plane perpendicular 
to / and containing P. Suppose, contrary to our assertion, that there 
are two distinct planes a and ft each perpendicular to / and each con- 
taining P. Then the intersection of a and {S is a line; call it m. 




Figure S-9 



Suppose I intersects m as in Figure 8-9. Then I intersects m in a 
point Q different from P. Why? Then a and /3 are each perpendicular 
to I at Q t But this is impossible. Why? Therefore I does not intersect m. 
Therefore there are two distinct points Q and K, as in Figure 8-10, 
such that I ± a at Q and I 1 £ at R. Then ARPQ has two right angles. 




Figure H-IQ 

But this is impossible. Why? Therefore there cannot be two distinct 
planes each perpendicular to I and each containing P. Hence there is at 
most one plane perpendicular to I and containing P. 



Definition 8.2 If A and B are distinct points, the unique 

plane that is perpendicular to AB at the midpoint of SB is 
called the perpendicular bisecting plane of AB. 



342 Perpendicularity and Paralleflwt In Space 



Chapter 8 



THEOREM 8.5 The perpendicular bisecting plane of a segment 
is the set of all points equidistant from the endpoinls of the segment. 



Let a be the perpendic- 
ular bisecting plane of A B and let 
C be the midpoint of A5. (See 
Figure 8-11.) Let P be a point. 
There are two things Lo prove, 

1. IfPCa J thcnAP = ¥R 
%, If AP = PB,thenf€a, 

You are asked to prove these two 
statements as exercises. 




Figun* SI 1 



THEOREM 8,6 Given two perpendicular lines, there is a unique 
line that is perpendicular to each of the given lines at their point of 
intersection. 



Proof: Let I and m be lines that are perpendicular to each other at P 
as in Figure 8-12. l^et a be the unique plane that is perpendicular to 
I at P, Let n be the unique line in a that is perpendicular to m at P. 
Then n is also perpendicular to L Why? Therefore there is at least one 
line perpendicular to both I and m at their point of intersection. 




Figure S-I2 

Suppose that it and n' are distinct lines each perpendicular to m 
and to I at P as in Figure 8-13. 



Figure S-13 




8,4 Other Perpendicularity Theorems 343 

Let a be the unique plane that is perpendicular to I at P. Then n, n', 
m are three distinct lines in a and n 1 m, n' _L m. Since this is impos- 
sible, there is at most one line perpendicular to both I and m it then- 
point of intersection. 

Since there is at least one line and at most one line that is perpen- 
dicular to / and in at P t it follows that there is one and only one line 
that is perpendicular to both I and m at their point of intersection. 

THEOREM 8.7 If two lines are perpendicular to the same plane, 
they are parallel. 



Let I and m be lines that are perpendicular to plane n\ as in 
Figure 8-14- If I = m, then l\,m and there is nothing more to prove. 
Suppose, then, that I ^ m. Then it can be shown that I and m are non- 
intersecting lines. (If / and m intersect at a point A not in a, then A 
and the two points where I and m cut ct are the vertices of a triangle 
with two right angles. If I and m intersect at a point A in or, then the 
plane containing I and m intersects a in a line n such diat n ll,ni m,} 




Figure S-14 

1 *-t F and Q be the points in which / and m, respectively, intersect 
a. Suppose, contrary to what we shall prove, that / and m are skew lines. 
I^t m* l>e the unique line through Q and parallel to I. Then / _L PQ 
and m* _ PQ. Why? Let t and & be die unique lines in a that are per- 
pendicular to r\) at F and at Q f respectively. Then I ± r and it fol- 
lows from Theorem 7.26 that m' _L ,v, Since m' J_ PQ and m* I 5, it 
follows that m' _L a. Let }i be the plane that contains m and »i' and 
let t be the line in which /? and a intersect. Then m _L £ m' J_ (, and 
the three distinct lines m f m', t are eoplanar. xSince this is impossible, 
it follows that I and m are not skew lines* Therefore they are coplanar 
nonintcrsecting lines, that is, they are parallel. 



344 Perpendicularity and Parallelism in Space 

THEOREM S.8 



Chapter 8 



If one of two distinct parallel lines is perpendicu- 
lar to a plane, then the other line is also perpendicular to that plane. 

Proof: Given two distinct parallel lines / and m and a plane « per- 
pendicular to l t let P and Q be the points in which / and m intersect tx t 

as suggested in Figure 8-15, Let r s PQ, let s be the unique line in a 
perpendicular to r at Q, and let mf be the unique line perpendicular 
to r and s at Q (see Theorem 8,6). It follows from Theorem 8.2 that 
m' is perpendicular to or., from Theorem 8.7 that m* is parallel to h, 
and from the Parallel Postulate that m' = m. Since m f is perpendicular 
to a, it follows that m is perpendicular to a. 

•I 




Figure S-15 

THEOREM 5,9 Given a plane and a point, there is a unique line 
containing the given point and perpendicular to the given plane. 

Proof: Let a plane a and a point P be given. We shall prove that there 
is a unique line through P and perpendicular to a. We divide the proof 
into four parts: two existence proofs 1 and 3 and two uniqueness proofs 
2 and 4. 

1. If P € a, we must prove that there is 
at least one line perpendicular to or 
at P. 

2. If P £ <*. w c must prove that there is 
at most one line perpendicular to at 
at P. 

3, If P g a, we must prove that there is 
at least one line perpendicular to ot 
and containing P. 

4, If P g a, we must prove that there is 
at most one line perpendicular to a 
and containing P. 



8,4 Other Perpendlculirity TTworems 346 

Proof of 1: Given a plane a and a point P in a as suggested in Figure 
8-16, let m and n be two lines in a that are perpendicular to each other 
at P. Then it follows from Theorem 8.6 that there is a unique line I that 
is perpendicular to both m and n at P. Then I -_ a at P. This proves that 
there is at least one line perpendicular to a at P. 




Figures- 1(1 

Proof of 2; Given a plane a and a point P in a, as suggested in Figure 
8-17, suppose contrary to what we shall prove, that m and n are dis- 
tinct lines and that each is perpendicular to a at P. Let /? be the plane 
that contains m and n and let I be the line in which /? intersects «. Then 
m l/,nl f, and the three lines I, m, n arc coplanar. This is impossible. 
Therefore there cannot he two distinct lines through P each perpen- 
dicular to a. 




Figure SI 



Proof of 3; Given a plane a and a point P not in a, as suggested in 
Figure 8-18, we shall prove that there is at least one line through P and 

perpendicular to a. Let Q be any point of a. If PQ J- a, there is noth- 
ing more to prove. Suppose, then, that FQ is not perpendicular to a. 



346 Ptrjwndicularity and Parallelism In Sf»c« 



Chapter 8 



Let I be the unique line such that IX a at Q. How do you know there 
is one and only one such line J? Let m he the unique line through P and 
parallel to /, Then it follows from Theorem 8.8 that m L <*. Therefore 
there is at least one line through P and perpendicular to a. 




Figure 848 



Proof of 4: Let a plane a and a point P not in a he $ven. If there 
were two distinct lines through F each perpendicular to a, then there 
would be a triangle with two right angles. Since this is impossible, it 
follows that there is at most one line through P and perpendicular to a. 

EXERCISES S.4 



1. Let four noncoplanar points A, B, C, D such that KB at "KC S6 AD, 
KB _L AC) AB _L A75, A7? JL ]S75 he given. Draw an appropriate figure 
and prove dial A BCD is not a right triangle. 

2. I ,et four noncoplanar points A , B, C, D such tliat AB _L AT7, KB JL AD, 

7VC _L AD be given. If A? is the bisector of L BAC, prove that / DAE 
is a right angle. 

3. Let t m, n be three distinct lines, not necessarily coplanar, and such that 
1 1| m and m |j n. Use Theorems 8.4, 8.7. and 8.8 to prove that I j| n. 

(This proves that parallelism for lines is a transitive relation.) 

4. Let A, B, C t D, E, F be distinct points such that A-B-C, AB = BC, 
AD = DC, AE = EC, AF = FC. Draw an appropriate figure and ex- 
plain why B f E, D, F are copianar points. 

5. Let segments AS and UD and three points E, W t G be given. If AT? and 
UD are perpendicular and intersect at E, if F-E-G and FT. L AB, docs 

it follow necessarily that FG is perpendicular to plane ABC? Draw an 
appropriate figure. 

6. \jet segments AB and C/5 and three points £. F, C be given. If AB and 
CD are perpendicular and intersect at E, if F-E-G, TG* X AT?, and 

FD _ CD, does it follow necessarily that FG is perpendicular to plane 
A BC ? Draw an appropriate fiu,i IE& 



8.5 Parallelism for Lines and Planes 347 

7. In part 2 of the proof of Theorem 8.4, how do wc know that the inter- 
section of a and ft is a line? 

8. In purl 2 of the proof of Theorem 8.4, how do we know that ml y at P? 

9. In part 3 of the proof of Theorem 8.4, how do we know that P lies in a? 

10. Given two distinct points A and B in a plane a describe the set of all 
points in a that arc eqtridistant from A and B. 

11. Givon two distinct points A and B describe the set of all points that are 
equidistant from A and B. 

12. Given a line / and the set § of all lines parallel to /, prove that if F is any 
point in space, then P lies on one and only one of the lines in $. 

13. Give a plane a and the set § of all lines perpendicular to a, prove that 
if P is any point in space, then P lies on one and only one of the lines 
ing, 

14. See the proof of Theorem 8.5. Prove that if P £ a y then AP = PB. 

15. Sec the proof of Theorem 8,5. Prove that if AP = PB, then P £ a. 
IG. Given A ABC and two points DandE such that BA 1 AB, UA 1 AC, 

EE 1 AB : EB 1 EC, prove that A E, A, B are coplanar points, 
17. See the proof of Theorem 8,9. In part 4 of this proof wc asserted that if 
there were two distinct lines through P each perpendicular to «, then 
there would be a triangle with two right angles. Draw an appropriate 
figure and prove this assertion. 

8.5 PARALLELISM FOR LINES AND PLANES 

In this section we shall investigate the properties of planes parallel 
to planes and of lines parallel to planes. Wc begin with two definitions. 

Definition 8.3 Two planes are parallel if their intersection 
is not a line. 

Definition 8,4 A line and a plane are parallel if their inter- 
section is not a point. 

Let us consider Definition 8.3 first, If a and ft arc planes, then the 

intersection of a and ft is (1) the null set, or (2) a line, or (3) a plane. 

If* n ft = t then a^=ft and a \\ft. 
If a n ft is a plane, then « H ft = a = ft and a || ft. 
If a n ft is a line, then a is not parallel to ft. 
Therefore a is parallel to ft if and only if 

a = ft or a = . 



348 Perpendicularity and Parallel*™ in Sface 



Chapter 8 



It follows immediately from Definition 8,3 that parallelism for planes 
is reflexive and symmetric. Later we shall see that parallelism for planes 
is also transitive, and therefore an equivalence relation. 

The fact that parallelism tor lines is an equivalence relation follows 
from Definition 7.1, and Theorems 8.4, 8.7, and 8.8. See Exercises 10, 
11, 12 of Exercises 7.6 and Exercise 3 of Exercises 8.4. 

Now let us consider Definition 8.4. Tf Ms a line and a a plane, then 
the intersection of I and of is (1) the null set, or (2) a point, or {3) a line. 
If I fl a = , then / 1 1 «. If / fi a is a line, then that line is t and 1 1 a. 
Ul n a is a point, then I and a are not parallel. Therefore I , a if and 
only if I C a or t H a = 0, 



THEOREM 8.10 If a plane intersects one of two distinct paral- 
lel hues but does not contain it, then it intersects the other line and 
does not contain it. 



Proof: (See Figure 8-19,) Let I and m be two distinct parallel lines. 
Let a be the plane that contains / and m, I^et /? be a plane that inter- 
sects one of the lines, say I, in a single point P. Now P lies in both ft 
and a. Since a contains / and $ does not contain /, it follows thai a =£ ft. 
Therefore the intersection of ft and a is a line; call it n. Since n lies in ft 
and I does not lie in ft t it follows that n is different from I. Also, m does 
not lie in ft. (Suppose m lies in ft. Then m lies in a and ft, and m = n. 
Since n is not parallel to l t it follows that m is not parallel to L Con- 
tradiction. Therefore m does not lie in ft,) 




Figure S-19 

Therefore m, n, I are distinct lines in plane a with I parallel to m 
and with n intersecting I in a single point. It follows from the Parallel 
Postulate that n intersects m in a single point; call it Q. (Suppose n 
does not intersect m. Then n and £ are two distinct lines through P and 
parallel to m. Contradiction.) It follows that ft contains Q and hence 
that ft intersects m in a single point 



8.5 Parallelism for Lints and Planes 349 

THEOREM S. II If a plane is parallel to one of two parallel lines, 
it is parallel to the other also- 

Proof: Assigned as an exercise, 

THEOREM 8.12 If a plane intersects two distinct parallel planes, 
the intersections are two distinct parallel lines. 

Proof: Let a and ft be two distinct parallel planes and y a plane that 
intersects both of them, as suggested in Figure 8-20. Then y is distinct 
from a and from fi t and it intersects each of them in a line. Let / and m 
be the lines in which y intersects a and ft t respectively; then I and m 
arc distinct coplanar lines which do not intersect. Why do they not 
intersect? Therefore I and m are distinct parallel lines. 




Figure 8-20 



THEOREM 8.13 If a, ft y are th ree distinct planes si ich that /? is 
parallel to y and such that a intersects /?, then a intersects y. 

Proof: (See Figure 8-2 L) Suppose that a, /?, y are distinct planes, that 
/? and y are parallel, and that a intersects 0, 




Figure 8-21 



350 Perpendicularity and Parallelism In Space 



ChaptBf 8 



Suppose, contrary to the assertion of the theorem, that a does not 
intersect y. Let m he the line in which a and /? intersect. Let P be a 
point of m; let I be a line through P that lies in a but not in ft and R 
a point of y. Let 8 be the plane containing I and R, Then 5 intersects $ 
in a line n distinct from m, and B intersects y in a line r. Then / and n 
are distinct coplanar lines through P and parallel to r. Since this con- 
tradicts the Parallel Postulate, it follows that a intersects y. 

THEOREM 8,14 If a line intersects one of two distinct parallel 
planes in a single point, then it intersects the other plane in a single 
point. 

Proof: Let a and ft be two distinct parallel planes and let I be a line 
that intersects a in a single point; call it F, (See Figure 8-22.) Let y be 
any plane that contains I, Then y and a arc distinct intersecting planes. 
Their intersection is a line; call it m. It follows from Theorem 8.13 and 
Theorem 8.12 that y intersects /? in a line; call it n. Then m and n arc 
distinct parallel lines that arc coplanar with I. Since / intersects m and 
is distinct from it, it follows from the Parallel Postulate that {intersects 
n and is distinct from it. Therefore the intersection of I and n is a single 
point; call it Q Since Q is a point of n, it is also a point of f$. Therefore 
Q is the unique point in which / and j8 intersect 



Figure 8-22 




THEOREM H. 15 If a line is parallel to one of two distinct parallel 
planes, it is parallel to the other plane. 

Proof: Assigned as an exercise. 

THEOREM 8 JO There is a unique plane that contains a given 
point and is parallel to a given plane. 



8.5 Parallelism for Line* and Plane* 351 

Proof: (We give only a plan for a proof.) Let a point P and a plane or 
be given. Divide the proof into four major parts. 

1. If P is in a, explain why there is at least one plane containing P 
and parallel to a. 

2. If P is in a, explain why there is only one plane containing P and 
parallel to or. 

3. If P is not in a, let / he the unique line through P and perpendicu- 
lar to a . Let ft be the unique plane that is perpendicular to I at P, 
Prove that a j /?, 

4. If P is not in a, prove that there is at most one plane through P 
and parallel to a. 

THEOREM 8.17 Given a point and a plane, then every line con- 
taining the given point and parallel to the given plane lies in the 
plane containing the given point and parallel to the given plane. 

Proof: (Plan only.) Let a point P and a plane a be given. Suppose first 
that P lies ma. Then a itself is the unique plane containing P and par- 
allel to «. Show that the assertion of the theorem is true in tins case. 

Suppose nest that P is not in a. Let /? be the unique plane contain- 
ing P and parallel to «. Then a=£fi. I.et I be any line through P and 
parallel to a, as suggested in Figure 8-23. Use Theorem 8.14 to com- 
plete the proof. 




Fi'gure S-2rs 



EXERCISES SJ 



L If a and j$ are planes such that or n fi = 0, is n f /J? Why? 

2. If / is a line and a is a plane and !n«= 0, is it possible that (a) t is 
parallel to a? (b) / is not parallel to a? Explain. 

3* Let a; /?, y be three distinct planes with a parallel to # and a not par- 
allel to y- From which theorem of this section may we conclude that $ 
is not parallel to y? 



352 Perpendicularity and Parallelism In 5p*c* Chapter 8 

4. Prove Theorem SJLt. 

5. Given a plane a and two distinct parallel lines I and m such that I is not 
parallel to a. From which theorem of this section may wc conclude that 

m is not parallel to a? 

6. See the proof of Theorem 8.13. Draw a figure like Figure 8-21. Add some 
segments and a label to the figure to suggest the plane & 

7. Prove Theorem 8.15. 



8.6 PARALLELISM AND PERPENDICULARITY 

hi tins section we define measure of a dihedral angle, right dihedral 
angle, and perpendicular planes. This section includes, as you might 
have guessed, several theorems on parallelism and perpendicularity for 
lines and planes. Wc begin by repeating a definition from Chapter 4, 



Definition 4.20 If two noncoplanar halfpknes have the 
same edge, then the union of these halfpknes and the line 
which is their common edge is a dihedral angle. The union of 
this common edge and either one of these two halfpknes is a 
face of the dihedral angle. The common edge is the edge of 
the dihedral angle. 



Figure 8-24 suggests two dihedral angles, one of them appearing to 
be larger than the other. What does "larger" as used here mean? In 
the case of (plane) angles wc defined larger in terms of measure. Since 
we do not have a measure for a dihedral angle, we develop one. 




Figure S-24 



Definition 8,5 The intersection of a dihedral angle and a 
plane perpendicular to its edge is a plane angle of the 
dihedral angle. 



8.6 Parallelism and Perpendicularity 353 



Figure $-25 suggests a dihedral angle A-BC-D, a plane a such that 
BC _i_ a, and L PQR, a plane angle of the dihedral angle A-BC-D. 




Figure s-as 

It seems plausible that all plane angles of a given dihedral angle 
should he congruent. This brings us to the next theorem. 

THEOREM 8. 18 Any two plane angles of a dihedral angle are 
congruent. 



Proof: Suppose that I ABC and LDEF 
are two plane angles of dihedral angle 
(2—Hl-J t as in Figure 8-26', We may suppose 
mat points have been picked in these plane 
angles and then labeled so that A and D lie 
in the plane GHI, C and F lie in plane IHJ, 
and 

AB = BC = DE = EF. 

It is easy to show that BCFE and BADE are 
parallelograms, that ACFD is a parallel- 
ogram, that AABCs A DBF, and hence 
IABC& LDEF, 




With Theorem 8.18 proved, it is now possible to define a measure 
for dihedral angles. 



Definition S. 6 The measure of a dihedral angle is ihe meas- 
ure of any one of its plane angles. 



Perpendicularity and Parallelism in Spact 



Chapter 8 



Definition 8. 7 A right dihedral angle is a dihedral angle 
whose measure is 9G> 



Definition 8.8 Two planes are perpendicular if their union 
is the union of four right dihedral angles. 



THEOREM 8. 19 If a line is perpendicular to a plane, then any 
plane containing the given line is perpendicular to the given plane. 

Proof: Assigned as an exercise, 

THEOREM 8.20 If a line is perpendicular to one of two parallel 

planes, then it is perpendicular to the other plane also, 

Proof: Assigned as an exercise. 

THEOREM 8.21 If two planes are perpendicular, then any line 
in one of the planes and perpendicular to their line of intersection 
is perpendicular to the other plane. 

Proof: Assigned as an exercise. 

TfTEOREM 8,22 If two distinct intersecting planes are perpen- 
dicular to a third plane, then their line of intersection is perpen- 
dicular to the third plane. 

Proof: Figure 8-27 suggests two distinct intersecting planes a and fi 
perpendicular to plane y. Let I = AB be the line of intersection of a 
and fii let EC and BD he lines in which m and ft respectively, inter- 
sect v. Let h be the line in a that is perpendicular to BC at B. 



Figure 8-27 




8.6 Parallelism and Perpendicularity 



Let ?2 be the line in /3 that is perpendicular to SD at B, It follows from 
Theorem 8,21 that h ± y and l 2 i y. The proof may be completed 
by showing that Ii ss-% a 1, and hence that / J. y< 



Definition &9 A segment, or ray, is perpendicular to a 
plane if the line which contains it is perpendicular to the 
plane, If a segment is perpendicular to a plane and otie end- 
point lies in the plane, then that segment is a perpendicular 
to the plane, and its endpoint in the plane is the foot of the 
perpendicular. 

Definition 8.10 If « is a plane and S is a set of points, then 
the projection of S on a is the set of all points (>, each of which 
is the foot of the perpendicular from some point of S, 



THEORFM 8.23 The projection of a line on a plane is either a 

line or a point. 

I-et t lie a line and a a plane. There are three cases to consider. 

1. /lies in cr. 

2. / is perpendicular to a. 

3- / is not perpendicular to a and does not lie in <*. 

PrcH>fofCase 1: If J lies in a t then every point of / is its own projection 
on a t and therefore J is its own projection on a. 

Proof of Case 2: If / is perpendicular to a at the point Q, then the pro- 
jection of every point of t on a is the point Q, and therefore die projec- 
tion of / on a is the point Q\ 

Proof of Cam 3: If / is not perpendicular to or and docs not lie in a T 
let A, B, C be three distinct points of / that are not in a, and let A', B', 
C be their respective projections on a. It follows from Definition 8. 10 

that AA' ± «, BB' J_ a, CC - a, and from Theorem 8J that AA' || 

BB\ BB' \ CC, AA' CC< Since / is not perpendicular to a and since 

A, B, C are distinct points, it follows that AA' r BB', CC* are distinct 

lines, T^t /3 be the plane that contains AA' and BB'. Then /3 contains 

the point C. Why? Since /? is parallel to S3' (Why?) and M' Is parallel 

to CC7, it follows from Theorem 8.11 that jS is parallel to £<?. Since fi 

contains C and is parallel to CC7, it follows that jS contains CC 1 . There- 



356 Perpendicularity and Parallelism In Space Chapter S 

fore C lies in fi. Let f be the line in which /? and a intersect. Since 
A', B\ C all lie in a and in fi, it follows that they all lie on V. If / in- 
tersects a t say in point D, then D is its own projection on a, and since 
D lies in both a and /•?, it follows that D lies on V , 

Note that I is determined by A and B, and that V is determined by 
A' and B*. We have shown that if P is any point of I distinct from A and 
B (like C and D in the preceding paragraph), then its projection F on a 
lies on t. 

Conversely, if Q' is any point of l\ then Q* is the projection of some 
point Q on (, If I and a intersect in {X then Q? is its own projection 
and Q s £7. If Q' is a point of P not on L then there is a unique per- 
pendicular to a at Q' ? and this perpendicular intersects ( in the point 
Q which has the point Q* as its projection on or. It follows that every 
point of / has some point of V as its projection, and that every point of 
r is the projection of some point on I. Therefore the projection of I on 
« is the line V, 

COROLLARY 8.23.1 The projection of a segment on a plane is 
either a point or a segment. 

Proof: Let s be a segment, or a plane, and $' the projection of s on a. 
If s is perpendicular to «, then *' is a point If s lies in a r then s' = s 
and hence s' is a segment. Let I be any line not in a and not perpen- 
dicular to a. Let A, B, Che three points of I such that A-B-C and let 
A\ R, C be their respective projections on a. Then it follows from 
Theorem 8.23 that A' } B\ C are collinear and from Theorem 8,7 that 
A'-R'-C. (IfA'-B'-C were not true, then we would have distinct par- 
allel lines that intersect,) Since A-B-C on I implies A'-B'-C, we say 
that betweermess for points is preserved in the projection from ion a, 
If s = AC and s' = WU, then B is between A and C if and only if ff 
is between A' and C. Therefore s' is the projection of s. 

THEOREM 8.24 Given a plane a and two distinct points A and 

B su ch tha t AB is parallel to a t if AW is the projection of XB on a, 
then A'W^AB. 

Frtxtf: Assigned as an exercise, 

THEOREM 8.25 Given parallel planes a and /? and A A RC in a, 
if A', B', C are the projections of A, B, C, respectively, on j5, 

then AARC^&A'B'C. 
Proof: Assigned as an exercise. 



8.6 Parallelism and Perpendicularity 

THEOREM 5.26 The shortest segment joining a given point not 
in a given plane to a point in the given plane is the perpendicular 
that joins the given point to its projection in the given plane. 

Roof: Assigned as an exercise. 

THEOREM 8.27 All segments that are perpendicular to each of 
two distinct parallel planes and have their endpoints in these planes 
have the same length. 

Proof: Assigned as an exercise. 



Definition 8.11 The distance between a point and a plane 
not containing it is the length of the perpendicular segment 
joining the given point to the given plane. 



Definition 8.12 The distance between two distinct parallel 
planes is the length of a segment that joins a point of one of 
the planes to a point of the other plane and is perpendicular 
to both of them. 



EXERCISES 8.6 



1. There is no definition of berweenness for half planes in this book. Write 
your own definition for it. Using your definition, prove that if 3C lt 3C 2 , 
X3 arc half planes with 3C Z between 0Ci and 3Ca, then the measure of 
the dihedral angle formed by 5C X and 3C 3 and their common edge is the 
sum of the measures of the dihedral angles formed by 3C j and 3C2 and 
their edge and by 3Ca and 3Cs and their edge. 

2. There are no definitions in this book for Ihc interior and the exterior of 
a dihedral angle. Write your own definitions for these terms. 

3. Draw a picture of a cube and label its vertices. How many right di- 
hedral angles are there each containing two faces of the cube? 

4. Given the labeled cube of Exercise 3, select one of the right dihedral 
angles and, using the vertices of the cube, identify two of its plane 
angles. 

5. See the proof of Theorem 8.18- Draw a figure like Figure 8*26 and mod- 
ify it to show the quadrilateral ACFD. Prove that ACFD is a rectangle. 



358 Perpendicularity and Parallelism In Spaca Chapter 8 

B Exercises 6-16 are concerned with projections on a plane. In Exercises 6-14, 
sketch one or more figures and he prepared to defend your answer. 

0, Is the projection of every triangle a triangle? 
7. Is the projection of every ray a ray? 
H. Is the projection of every point a point? 
9. Is the projection of every segment a segment? 

10, Ls the projection of every line a line? 

1 1. Is there an angle whose projection is a line? \ ray? A segment? A point? 
An angle? 

12. ls there an acute angle whose projection is an obtuse angle? 

13, Is there a right angle whose projection is an acute angle? An obtuse an- 
gle? A right angle? 

14. Is there a segment that is shorter than its projection? linger? 

15, If an edge of a cube is perpendicular to a plane, describe the projection 
of the cube on the plane. 1 5 raw a sketch, 

10, challenge problem. If a diagonal of a cuIhj is perpendicular to a 
plane, draw a sketch of the projection of the cube on the plane. 

■ In Exercises 17-23, draw an appropriate figure and prove the theorem, 

17. Theorem 8,19 21. Theorem 8.25 

18. Theorem 8.2G 22. Theorem 8-26 

19. Theorem 8.21 S3. Theorem 8.27 

20. Theorem S-24 

24. In the proof of Theorem 8,22, show that h = fe = '■ 

25. Let A-BC-D be a right dihedral angle with M 1 B€, EC 1 CD, 
AB = 3 v#, BC = 5, CD = 12, Prove that AD = 14. 

26. The figure shows four noncoplanar points A, B, Q D such that 

AB _L AC, AB J. AT5. XC I AD, AB = AC = AD. 
Find the sum of the measures of I BDA, LBDC, L CDA. 




Chapter Summary 359 

in Exercises 27-34, determine if the given statement is true or false. (An 
if-then statement is false if there are one or more instances in which the 

if-part is Lrue and the then-part false.) 

27. If a line is perpendicular to two distinct intersecting lines at their point 
of intersection, then it is perpendicular to the plane that contains the 
intersecting lines. 

28* If the intersection of a plane and a dihedral angle is an angle., then that 
angle is a plane angle of the dihedral angle. 

29. If Z and n arc distinct intersecting lines and if I is parallel to plane « t 
then n is not parallel to plane a. 

30. If T and n are distinct parallel lines intersecting (hut not contained in) 
two distinct parallel planes, then the planes cut off segments of equal 
length on the two lines. 

3t< If two planes are perpendicular to the same line, they arc parallel. 

32. If two Kltet are parallel to the same plane, they are parallel. 

33* Given a plane a, the set of all points each of which is at a distance of 
5 from a is the union of two planes each parallel to a. 

34. If a is a plane and F is a point, there is one and only one plane contain- 
ing V ami parallel to a. 



CHAPTER SUMMARY 

We l>egan the chapter with some suggestions for drawing figures to 
represent geometrical figures that do not lie in a single plane. We used fig- 
ures throughout to help you visualize the relationships of lines and planes 

in space. 

Definitions of the following expressions were included. 

A LINK AND A PLANE .ARE PERPENDICULAR 

THE PERPENDICULAR BISECTING PLANE OF A SEGMENT 

PARALLEL PLANES 

A LINE AND A PLANE ARE PARALLEL 

A PLANE ANGLE OF A DIHEDRAL ANGLE 

THE MEASURE OF A DIHEDRAL ANGLE 

A RIGHT DIHEDRAL ANGLE 

PERPENDICULAR PLANES 

A SEGMENT (RAY) PERPENDICULAR TO A PLANE 

THE PROJECTION OF A SET ON A PLANE 

THE DISTANCE BETWEEN A POINT AND A PLANE 

THE DISTANCE BETWEEN TWO PARALLEL PLANES 

There are 27 theorems in this chapter. It is suggested that you write 

out these theorems and draw appropriate figures for them. 



360 Perpendicularity and Parallelism In Space Chapter 8 

REVIEW EXERCISES 

■ In Exercises 1-15, copy and complete the given statement to obtain a the- 
orem of this chapter. 

1. If a line is perpendicular to each of two distinct intersecting lines at 
their point of intersection, tl icn 3 

2. Given a line and a point, there is a unique plane perpendicular [T]. 

3. The perpendicular bisecting plane of a segment is the set of all points 
each oi' which is |TJ, 

4. If two lines are perpendicular to the same plane, they are ]TJ, 

5. If one of two distinct parallel lines is perpendicular to a plane, then \T\. 

6. Given a plane and a point, there is a unique line containing the given 
point and |T|. 

7. If a plane intersects one of two distinct parallel lines but does not con- 
tain ft, then it \TJ and does not contain it. 

8. If a plane intersects each of Lwo distinct, parallel planes, the intersec- 
tions arc [7] lines. 

9. If a, /?, y are three distinct planes such that a is parallel to /? and such 
that n intersects y t then [?]- 

10. If a line intersects one of two distinct parallel planes in a single point, 
thenQ]. 

11. If a line is parallel to one of two distinct parallel planes, then \T}. 

12. Given a plane and a point, there is a unique plane that contains the 
given point and [?]. 

13. Any two plane angles of a dihedral angle are [7}* 

14. If a line is perpendicular to a plane, then every plane containing that 
line Is Q]. 

15. If two distinct intersecting planes are perpendicular to a third plane, 
then their line of intersection is QJ 

■ In Exercises 16-23, some lines or planes are described. In each exercise, 
state whether or not they must be parallel to each other, 

16. Lines through a given point parallel to a given line. 

17. Lines perpendicular to a given plane. 
IS. Lines perpendicular to a given line. 

19. Lines parallel to a given plane. 

20. Planes parallel to a given plane. 

21. Planes perpendicular to a given plane. 

22. Planes perpendicular to a given line. 

23. Planes parallel to a given lino. 



Review Exercises 361 
In Exercises 24-30, write a plan for a proof of the given statement. 

24. If TF is the projection of A£ on a plane, then A'B' < AB. 

25. If A'B' is the projection of AB on a plane a and if A'B' — AB, then 
AB | a. 

26. If a and B are distinct parallel planes, if parallel lines m and n are in /?, 

and if lines m' and n f are the projections of m and n, respectively, on « T 
then m' is parallel to n'„ 

27. If « and /J are distinct parallel planes, if $ and f are parallel and con- 
gruent segments in t and if s' and f arc the projections of * and t, re- 
spectively, on a, then $' and f are congruent. 

28. If A$ I ifi?', AC 1 1 A 7 ?, and if A, B, C are noneollincar points, then 
plane ABC and plane A'B'C are parallel. 

29. If a and /? are distinct parallel planes and if I and m are distinct parallel 
lines that intersect ft in points Li and Mi, respectively, and /? in points 
J-2 and I4& respectively, then LiAfiJUgLi is a parallelogram. 

30. If every plane containing a given line is perpendicular to a given plane, 
then the given line is perpendicular to the given plane. 




er 



The Public Archives of Canada 



Area 

and the 

Pythagorean Theorem 



9.1 INTRODUCTION 

You are familiar with the idea of area and have computed the areas 
enclosed by figures such as triangles, sqtiaras, and circles. As you know, 
areas are usually computed using numbers which arc lengths or dis- 
tances. You also know that a distance function is based on some seg- 
ment as the unit of distance. In informal geometry, we usually combine 
a number and a word in expressing a distance, for example, 4.5 ft. In 
formal geometry, we frerjuendy use the number by itself, omitting the 
unit when the distance function is understood. Tn our formal develop- 
ment of the area concept wc shall suppose that a distance function is 
given. Then there will be just one area function- 
In this chapter we adopt some postulates for area based on our ex- 
periences in computing areas and on our ideas about areas in informal 
geometry. These postulates include the formula for computing the area 
enclosed by a rectangle as well as statements of general properties that 
are useful in proving theorems about area. We also develop formulas 
for computing the areas enclosed by figures such as triangles and trape- 
zoids, and sve use area as a tool in proving the Pythagorean Theorem. 



364 Area and the Pythagorean Theorem 



Chapter 9 



9,2 AREA IDEAS 

Let us suppose that a unit segment for distance measure is given 
and that a square of side length I is given. 
We call this square the unit square. (See Fig- 
ure 9-1.) Just as distances and lengths are 
numbers associated with sets of points (for 
example, the length AH is associated with 
AH), areas are numbers associated with sets 
of points. The distance function matches a 
positive number with each segment. Simi- 
larly, the area function matches a positive 
number with each rectangle and also with 
many other simple figures, 

We shall agree that the area of the unit square is J , In practical ap- 
plications there is a system of areas for each system of distances. If the 
side length of the unit square is 1 ft., then the area of the unit square is 
1 sq. ft. If the side length of the unit square is 1 cm,, then the area of 
the unit square is 1 sq, cm. (See Figure 9-2.) 



F%v«B-i 



1 aq. Lin. 



Figure 9-2 



1 so. in, 



When we speak of the area of the unit square as 1 , we are thinking 
of 1 as the measure of the set that includes afl of the points in the plane 
of the square that are inside of or on the 
square. (See Figure 9-3.) Strictly speaking, 
it would be better to speak of the area of die 
unit square region. We shall follow custom, 
however, and call it the area of the unit 
square. Similarly, we talk about areas of rec- 
tangles, triangles, and circles when we really 
mean the areas of the regions which are made 
up of these figures including their interiors. 
(See Figure 9-4.) 

When we say the "area of a triangle," for example, we mean the 
"area of the triangular region." 



Fljjure 9-3 



9.2 Area Ideas 



365 



Definition 9.1 A polygonal region is a triangular region, or 
it is the union of a finite numher (two or more) of triangular 
regions such that the intersection of every two of them is 
the null set, or a vertex of each of them, or a side of each of 
them. 




Polygonal 
region 



Union of 

triangular 

run ions 

Figure 9-4 



You may already know how 
to compute the area of a rec- 
tangle. If its sides are of length 
a and b, its area is ah, (See Fig- 
ure 9-5,) In Section 9.3 we 
adopt this formula for the area 
of a rectangle as a postulate. 
Let us see if it is really a reason- 
able and basic assumption to 




ure9-5 



Figure 9-6 suggests our unit square and a rectangle with sides of 
lengths 3 and 4. The rectangular region is made up of 3 * 4 = 12 square 
regions each of the same Size as the unit square region. So the area of 
the rectangle should be 3 • 4, or 12. Similarly if a and b are any two 
natural numbers (that is, positive whole numbers), then the area of a 
rectangle which is a by b should be ab. 



□ 



Figure 8-6 



I 



366 Area and the Pythagorean Theorem 



Chapter 9 



Figure 9-7 suggests our unit square and another square with sides 
of length ^, Let x denote the area of the square with sides of length L 
Since it lakes 9 squares of this size (3 rows with 
3 squares in each row as suggested in the figure) 
to fill up the unit square, you can see that the 
area of the unit square should be 9 times the 
area x of the ^ by y square. Therefore 
9x= 1 



□ 



* — 1 
x -$. 



l 



Figure 17 



It appears, then, that the area formula, 

S = ah, 

is a reasonable formula for our ^ by j- square. Taking as| and h = -k 
we have 



1 



the area of a ^ by ^ square. Similarly, if n is any natural number, then 
area of a — by — square is — • — , or -=r. 



n 

Figure 9-8 shows a rectan- 
gle which is ^ by |. It is easy to 
see from die figure that the area 
of the rectangle should be 2 • 5, 
or 1 0, times the area of a } by 4- 
square; hence 

Therefore the formula tS as ah is appropriate for a rectangle which is 
| by |, Similarly, if a and b arc any two positive rational number*, then 
the area S of a rectangle which is a by b should be given by the formula 
S = ah. For example, if 



-_ 
i 



Ekpre 9-8 



then 






6 = |, 

I, _ 3 _ 24 
" — 5 ~ 4I>* 



and if R is a rectangle which is a by £>, its area S is the sum of the areas 
75 • 24 = 1800 Utde squares, 

each of which is 4 ' by -£$, Hence 

Since l|go = |=Jf 4=(lM). 

we see that the formula S = ah holds, in this case. 



9,2 Area Ideas 

Let R be a rectangle which is a by b and suppose that either a or b t 
perhaps both, is an irrational number. Then, using the idea that an 
irrational number can be squeezed between two rational numbers that 
arc as close together as you desire, we can make it seem reasonable 
that the area S of R is again given by the formula S = ah, Figure 9-9 
shows a rectangle R which is 1 by yf%, (Recall that \/2 is not rational 
and that i/2 can be represented by the nonrepeating decimal 
1.41421 . . . .where the three dots indicate an infinite sequence of 
digits.) Thus 

L4< \/2<1.5> 

where 1 .4 and 1.5 are rational numbers, Now a rectangle 1 by 1.4 has 
area 1.4, a rectangle 1 by L5 has area 1.5, and therefore the area S 
of the 1 by \*2 rectangle R is somewhere l>ctween 1.4 and 1 .5, that \s t 

1.4 < S < 1.5. 



367 




Figure 9-9 

Continuing in this way* we can show that S is between 1.41 and 
1.42, between 1.414 and 1.415, and so on. If you consider the 
sequence of areas of rectangles 1.4, 1.41, 1.414, . . . , you see that the 
sequence is increasing and the rational numbers in the sequence are 
getting closer and closer to \/2 through numbers that are less than \/2, 
Similarly, if you consider the sequence of areas of rectangles 1.5, 1.42, 
1*415, . * . , you see that the sequence is decreasing and the rational 
numbers in the sequence are getting closer and closer to y/2, through 
muni tens that are greater than y/2. It seems reasonable, then, to con- 
clude that the area S of the 1 by \/2 rectangle R is given by 

Ssl'yfs y^. 

The foregoing discussion may be summarized as follows. If the 
sides of a rectangle are rational numbers, the rectangular region can 
be either subdivided into a finite number of unit square regions and the 
area obtained by counting, or subdivided into a finite number of square 

regions each — by — for some integer n > 2 and then counting and 



368 Area and the Pythagorean Theorem Chapter 9 

multiplying byi In both cases the result S is the same as that obtained 

by using the formula 5 = ah, where a and h are the lengths of two 
adjacent sides of die rectangle. If the side lengths are not rational, then 
we cannot subdivide the rectangular region into square regions with a 
rational side length and find the area by counting, or by counting and 

multiplying by -^- for some natural number a. We can, however, 
n 2 

approximate the side lengths using rational numbers and consider the 
limit of the sequence of areas of the rectangles whose sides have 
lengths that are rational numbers. It can be proved that this limit exists 
and is equal to the product ah* In this book we adopt the formula S=al> 
for the area of a rectangle as a postulate. 

The ideas used to develop the formula S" = ah for the area of a 
rectangle are, of course, the same ideas with which you are familiar 
from your past experiences in informal geometry. They include the 
following: (1) every rectangle has an area, (2) congruent rectangles 
have the same area> and (3) area is additive in die sense that the area 
of a figure is the sum of the areas of the parts of the figure. Similar 
ideas may be used in developing area formulas for triangles, parallelo- 
grams, trapezoids, and so on. In Section 9.3 we state carefully our basic 
ideas about area- We call these statements die Area Postulates and use 
them to develop several area formulas. 



9.3 AREA POSTULATES 

Our formal development of area in this chapter is limited to areas 
of polygons and is based on the following postulates. 

POSTULATE 27 (Area Existence Postulate) Every polygon 
has an area and that area is a (unique) positive nuxnlier. 

POSTULATE 81 (Rectangle Area Postulate) If two adjacent 
sides of a rectangle are of lengths a and b t then the area S of the rectan- 
gle is given by the formula S = ah. 

POSTULATE 29 (Area Congruence Postulate) Congruent poly- 
gons have equal areas, 

POSTULATE 30 (Area Addition Postulate) If a polygonal re- 
gion is partitioned into a finite number of polygonal subregions by a 
finite number of segments (called boundary segments) such that no 
two subregions have points in common except for points on the bound- 
ary segments, then the area of the region is the sum of the areas of the 



9.3 Ares Postulates 



169 



Notation, If 3 is a figure, we frequently use |ff | to denote the area of 
?. For example, &ABC\ denotes the area of A ABC. The area of quad- 
rilateral A BCD may he denoted by | quadrilateral ABCD\ or by \ABCD\ 
if it is clear that ABCD is a quadrilateral. 

Example J Figure 9-10 suggests a polygonal region ABCDEFG, Let 
us call this region (P. Segments CO and CP partition, or separate, (P 
into three subregions (Pi, (P2, (P3, where (Pi, is the union of quadrilat- 
eral ABCC and its interior, <S>z is the union of A CFG and its interior, 
and (P3 is the union of quadrilateral CDEF and its interior. According 
to Postulate 30 } the Area Addition Postulate, the area of (P is the sum of 
the areas of 6*1, (P & and (P3, that is, 

|<P| = ((P^ + |<Pa| + |(P 3 |. 

H C 




Example 2 Figure 9- 1 1 suggests a figure consisting of a square A BCD 
with its diagonals, AC and DB t and an adjoining isosceles right triangle, 
ABFC t with legs of length I, Let S denote the area of &BFC, S' the 
area of square ABCD, and S" the area of pentagon ABFCD. FindS, 
S\ S" using the Area Postulates. 




Figure U-ll 

Solution: It is easy to show that ABFC = A BEC SO that the triangles 
have equal areas and thai BFCE is a rectangle, actually a unit square. 
Then 

1 = \BFCE\ 
= \&BFC\ + \ABEC\ 
= J auk; + \ABFC\ 



by the Rectangle Area Postulate 
by the Area Addition Postulate 
by the Area Congruence Postulate 



370 Area and the Pythagorean Theorem Chapter 9 

Therefore 2S = 1 and S = j. Since square ABCD is partitioned by its 
diagonals into four triangles, each congruent to A BFC, it follows from 
the Area Addition and Congruence Postulates that S' = 4S and hence 
that 

S' = 4 • j = 2. 

From the Area Addition Postulate it follows that 

S" = S' + S = 2J. 

Another way to find S' is as follows. Since ABbXl = ABEC, we 
have 

BE = EC=l and mlBEC = 9Q. 

Then it follows from the Pythagorean Theorem, which is proved in 
Section 9.5, that 

(BC)2 = 12 + 12 = 2 and BC = \M. 

Finally, it follows from the rectangle area formula that 

S' = y/2 • \/£ = 2- 

EXERCISES U 

■ In Exercises 1-8, given the lengtta a and 6 of two adjacent sides of a rectan- 
gle, find the area of the rectangle. 

L a = 2,5, b = 3.3 5« = £ b = 4 

2.a = % t b = i 0, a = 7, b = 3yl" 

3.as|^s| 7, a = 4t/2, 6 = syi 

4 a = 2,32, 6 = 1.77 8, a= Vo y b = 5^ 

9. (An informal geometry exercise.) The length of a side of a square is 

10 yd. What is the area of the square in square yards? In square feet? 
10, (An informal geometry exercise.) ]f the dimensions of a rectangular 

field are 40 rods by 80 rods* how many acres of land does it contain? 

(An acre contains 43,360 sq. ft. and 1 rod is 16| ft. long.) 
1L (An informal geometry exervim.) How many feet of fence would it 

take to enclose the field in Exercise 10? 

12. (An informal geometry exercise.) The dimensions of a football field 
(including the end zones) are 53} yd, wide by 120 yd long. Would an 
acre of land be sufficient on which to lay out a football field? (See Ex- 
ercise 10,) 

13. If the perimeter of a square is 4H, what is its area? 

14* If the perimeter of a square is 64 \/2. what is its area? 



9,3 Area Postulates 



371 



15- The length of one side of a rectangle is 3 1 hues the length of an adjacent 
side. If the perimeter of the rectangle is 5fi, what is its area? 

16. The length of one side of u rectangle is 5 more than twiee the length of 
an adjacent side. If the perimeter of the rectangle is 70, what is its area? 

17. challenck pkoblem. The length of a rectangle is 4 more than twice 
its width. If the area of the rectangle is 48, what is its perimeter? 

IS. The width of a rectangle is | of its length. If the area of the rectangle 
is 360, find its length and width. 

U). (An informal geometry exercise.) How many squares 1 in. on a side 
are contained in a square 1 yd. on a side? 

20. (An informal geometry exercise.) How many 9 in. X 9 in. tiles would 
it take to cover a 12 \ ft. X 9 ft. rectangular floor if wc assume that there 
is no waste in cutting the tile? 

21. If S is the area of an x by x square and S' is the area of a 2.v by 2.v square, 
prove that S" = 43, 

22. If S is the area of a right triangle with legs of lengths a and b t and 5' is 
the area of an a by b rectangle, prove that S* = 2S. 

23. If S is the area of a right triangle with legs of lengths a and b t and 5' is 
the area of a right triangle with legs of lengths 2a and 2b, prove S' = 4S. 

24. The figure shows two coplanar equi- 
lateral triangles, AABC and AACP, 
with AC = 1 . The feet of the perpen- 
diculars from A and D to EC are E 
and F, respectively. Prove that ABCD 
is a parallelogram and that AEFD is a 
rectangle. 

25. Complete the proof of the following theorem. 




THEOREM The area of a right triangle is one-half the product of 
the lengths of its legs. 

restatkmkst: 

Given: A right triangle A ABC with 
the right angle at C, AC=b 7 
and DC = a. 



_ i 



Pram. |AAJ9C| = 




Proof: Let C be the point of intersection of the line through A and 
parallel to OB and the line through B and parallel to XU, (See the figure,) 
Why must these two lines intersect? YVhy is quadrilateral ACBC a 
rectangle? 

Complete the proof by showing that AABC ^ ABAC and that 
| A ABC = \ab. 



372 Ar«i And the Pythagorean Theorem 



Chapter 9 



9.4 AREA FORMULAS 

Figure 9-12 shows a parallelogram ABCD with points £ and F the 
feet of the perpendiculars from D and C, respectively, to line AB. 




Figure 9-12 

The following definition has two parts. In (1) we define base and 
altitude, thought of as segments. In (2) wc define base arid altitude. 
thought of as numbers (lengths of segments, or distances). (Compare 
the corresponding definitions for triangles on page 265.) 



Definition 9.2 

1 . Any side of a parallelogram is a base of that parallelogram. 
Given a base of a parallelogram, a segment perpendicular 
to that base, with one endpoint on the line containing the 
base and the other endpoint on the line containing the 
opposite side, is an altitude of the parallelogram corre- 
sponding to that base, 

2, The length of any side of a parallelogram is a base of that 
parallelogram. The distance between the parallel lines con- 
taining that side and the side that is opposite to it is the cor- 
responding altitude. Of height 



In Figure 9-12 we call h the altitude, or height, corresponding to 

the base AB. The number h is the distance between the parallel lines 

A3 and DC. Similarly, the distance between the parallel lines AD and 

BC is the altitude which corresponds to the base AD, 

Let S denote the area of parallelogram ABCD in Figure 9-12, let 
AB = ft, and let CF s ft, Wc proceed to prove that the area S of the 
parallelogram is bh. 

THEOREM 9.1 If h is a base of a parallelogram and if h is the 

corresponding height, then the area S of the parallelogram is given 
by the formula $ = bh. 



9.4 Area Formulas 



373 



talement 

L £DAE=& /CBF 

2. lAEDj= IBFC 

3. Z5E ss CF 

4. AAED m ABFC 

5. \AAED\ = \ABF€\ 

a \AFCD = S + |ABrcj 

7. \AFCD\ = S+ |AA£D' 

8. S= |AFCD| - |AAED| 

9. |£FC£>| + AAED\ = 
\AFCD\ 

in. OC = b 

1L |EFCD| = fc/i 

12. fefi + |AAFD| = \AFCD\ 

13. /?/i = |AFCD| - \AAED\ 

14. S = Mi 



Reason 

1. Why? 

2. Why? 

3. Why? 

4. S.A.A, Theorem 

5. Why? 
8. Why? 

7. Why? 

8. Addition Property of Equal- 
ity 

9. Why? 

10. Why? 

11. Why? 

12. Steps 9, 11; substitution 

13. Why? 

14 Steps 8, 13; substitution 



In the remainder of this section we prove two more theorems re- 
area formulas. 



THEOREM 9.2 If b is a base of a triangle and if h is the corre- 
sponding altitude, then die area S of the triangle is given by the 
formula S = ^bh. 

Proof: (See Figure 9- 13 . ) Let A A3 C be given . Let 55 be the ray with 

endpoint B and parallel to AC. Let CE be the ray with endpoint C and 

parallel to AB. Let F be the point of intersection of bB and CE. Then 
ABFC is a parallelogram and A ABC SB A FOB. 




Figure 9-13 



374 Area and the Pythagorean Theorem Chapter 9 

Now b m A3 and h is the distance from C to A$> and so /i is also the 

distance between CE and S3, It follows from Theorem 9.1 that 

ABFC\ = bh> But 

I A ABC] = \AFCB\ (Why?) 

and 

|AABC| + |AFCB| = \ABFC\ (Why?). 
Therefore 

S + S = 6Ji, 2S = &/», and S = Jbfc. 

THEOREM 9.3 If 6i and b 2 are the lengths of the parallel sides 
of a trapezoid and if h is the distance between the lines that con- 
tain these parallel sides, then the area S of the trapezoid is given by 
die formula 

S s 4(&i + b*)h. 

Proof: Let trapezoid ABCD t with parallel sides AH and CD, be given 
as in Figure 9-14. 



h 



/ 






A 

Figure 9-14 






______ — . , 



The lengths of these sides, or bases, are b\ and h% % and the distance 

between AB and CD is h. The distance to is called the height. Let E 

be the point on opp BA such that BE = Ify. Let F be the point on 

opp CD such that CF m b%. Then AEFD is a parallelogram (Why?) and 
its area is (bj + & 2 )n. We shall show that 

A BCD «— * FCBF 

is a congruence, Then it will follow that 

|ABCD| + \FCBE\ = \AEFD\ 

S + S = (hi + &_)fc 

S = •£(&! 4- &)fc. 



9,4 Area For mulit 375 

To see that ABCD «— * FCBE is a congruence, note that 

(1) LBAB = lEFC (5) XB E PC 

(2) ZABC a ZFCB (6) B'C <= GB 

(3) IBCD=* ICBE (7)C35 = B£ 

(4) LCDA = ABEF (8) DA sj EF 

This completes the proof. 

There are area formulas for other special polygons, but we shall not 
develop them here. In Section 9.5 we use area formulas us tools in prov- 
ing the Pythagorean Theorem. 

EXERCISES 9.4 

I* If the area of a triangle is 40 and a base is S.. find the corresponding 

altitude, 

2. If an altitude of a triangle is 12 and the area of the triangle is 62, find the 
base corresponding to the given altitude. 

3. The lengths of the legs ol a right triangle are 17 and 9. Find the area of 
the triangle. 

4* Find the area of an isosceles right triangle if the length of one of its legs 
is 15, 

5. Find the area of a rhombus if its height is 7j and the length of one of its 
sides is 15^. 

■ Exercises 6-8 are informal geometry exercises. 

6. In the figure below, lines m and n are parallel and the distance between 
them is 5 ft. P and Q are points on m, and R and S are points on n such 
that FQ = RS = 7 ft. If the distance from P to R k 10 miles, find 
PQRS. 

P Q 
4 — • ♦ 7/ btn 




8 



-M 



7. In the situation of Exercise 6, find FQHS if PR = 1 mile. 
& In the situation of Exerd.se 6, find \PQRS\ if PR = 10 ft, 

9* If the area of a triangle is S and its altitude is h, express the base b in 
terms of S and h. 

10. The area of a parallelogram is 50.4. Find its base if the altitude is 4.2. 



376 Area and the Pythagorean Theorem Chapter 9 

11. Find the area of a trapezoid if its height is 7 and the lengths of the par- 
allel bases are 9 and 15* 

12. Use the formula for the area of a trapezoid and express h in terms of 
5, hi, and b& 

13. If the area of a trapezoid ( s 128 and the lengths of the parallel bases are 
7 and 9, find its height, 

14. The area of a trapezoid is 147 and the lengLh of one of i ts parallel bases 
is 18. If the height is 7. find line length of the other base. 

15. ABCD is a parallelogram with AB = 25. If \ABCD\ = 375 and P is a 
point such that C-P-D, find | AAPB|. 

Ifl. An informal geometry exercise.) A plot of land in the shape of a 
trapezoid has bases of lengths 121 yd. and 242 yd. If the distance be- 
tween the bases is 300 yd., how many acres of land are contained in the 
plot? (One acre contains 43,560 sq. ft) 

17. Two triangles have the same base and their areas differ by 42, If the 
altitude of the larger triangle is 6 more than the altitude of the smaller 
one, find the length of the common base. 

18. Id the figure, ABCD is a trapezoid with parallel liases AB and CD. If 
AB m fej, CD = h z , and the altitude is h, draw ED and use the Area 
Postulates and the area formula for a triangle to prove that 

\ABCD\ = ^(fcx + /ia). 




19. In the figure, AM is a median of AABC. Prove that 

lAAJMfj = |AACMj. 
Is AABM r- A ACM in every case? Iu any case? 




The area of a square is equal to the area of a parallelogram . If the base 
of the parallelogram is 32 and its height is 18, find the perimeter of the 

square. 



9.4 Ar»a Formulas 377 



In each of Exercises 21-29, find, on the basis of the given data, the area or 
areas indicated, 

21, X3 I Be* BC = 8, 23, ABCD is a square, 

DE = 6. f AAfiCJ = [T]. E is the midpoint of HC 

The perimeter of ABCD is 48, 

A D IAACE =m. 




22. D-G-F r DF = 6, DE a 6.5, 
GE = 5. ADEF] = {T\. 





26. ABCD is a parallelogram, E is 
the midpoint of C~D r and 
|A£CDj =33. 
JAA£D] =[?] 
jAA£E| =0] 
|ASCE| = fl 



23, AH = CD = 8, BC = DA = 6, 
DE b 5. AABCD\ = JTj. 






27. A£CD is a square; K, F, G, // 
are midpoints of the sides; 
jABCD, = 196, \EFGH\ = [?]. 



ABCD is a parallelogram, EF ± 
M, £F =3, BC= 8, CD = 4. 

|AA/*CD| = [2]. 





378 Area and the Pythagorean Theorem 



Chapter 9 



2H. EFGH is a square of side length 6; ABCD is a square of sick length 12; 
E, F s C, It are interior points of ABCD. The area of the shaded region 




29. B and £ arc collinear with A and F ( K and 
H are collinear with L and (». C and / are 
collinear with B and /C, D and J are col- 
linear with E and H, AFCL is a rectangle 
with area 46, BCDE b ;i iert,mgjB with 
area 8, BCDE and jfKHJ are congruent 
rcctangjes. ABCDEFGHIJKL\ = [J 



fl 



J AT 



J H 



30. Using the Area Postulates and the theorem of Exercise 25 in Exercises 
9.3, derive the tri angle area formula for any triangle. 

31. (See Figure 9-12.) Let b\ = AD and let A, be the distance between 
AD and BC. Is £?i/»i = hh? Justify your answer. 

32. Would any of the statements in the proof of Theorem 9,1 be different 
if point £ were between A and B? Draw a figure for this case. 



33. See the proof of Theorem 9.2. Prove that Q? and BD are not parallel 
lines. 

34 In the proof of Theorem 9.2, prove that A ABC = AFCB. 

35. Recall that a parallelogram is a special trapezoid. Show that the formula 
for the area of a trapezoid simplifies to the formula for the area of a 
parallelogram if the trapezoid is a parallelogram. 

36. A triangle might be thought of as a quadrilateral in which one of the 
bases has shrunk to a point. What number does j[hi + b%)h approach 
if h t and h are fixed and h% gets closer and closer to 0? 

37. challenge problem. Prove that the area of the triangle determined 
by the midpoints of the sides of a given triangle is one-fourth the area 
of the given triangle, 



9.5 Pythagorean Theorem 379 



9.5 PYTHAGOREAN THEOREM 

In this section we state and prove the Pythagorean Theorem and 
its converse. Perhaps no other theorem In mathematics has inspired 
more mathematicians and nonmathematicians alike to find original 
proofs as has this one. There are over 370 known proofs of this the- 
orem of which over 255 employ the use of areas. The first proof of this 
theorem has been attributed by many historians to Pythagoras (582- 
501 Ji.c), a Greek mathematician and philosopher, although the prac- 
tical application of the theorem was known many years before his 
time. It is not known which proof Pythagoras gave, but in the Exercises 
at the end of this section is an outline of a proof which appeared altoul 
300 b.c. in the first of the thirteen books of Euclid's Elements. 

Perhaps the most unique proof of the Pythagorean Theorem was 
devised by a 16-year-old schoolgirl, Miss Ann Condi t, of South Bend, 
Indiana, in 1938. The proof devised by Miss Condit and an original 
proof devised in 1939 by Mr. Joseph Zelson, an 18-year-old junior in 
West Philadelphia, Pennsylvania High School, show that high school 
students are capable of original deductive reasoning. For a list of proofs 
of the Pythagorean Theorem, including the proofs by Miss Condit and 
Mr. Zelson, see The Pythagorean Proposition by Elisha Scott Loomis, 
published by the National Council of Teachers of Mathematics, 1968. 

THEOREM 9.4 (The Pythagorean Theorem) If a, b, c are the 

lengths of the sides of a right triangle A ABC, with c = AB t the 
length of the hypotenuse, then 

a 2 + &>=<& 



Let BC = a, CA = h. Let PQRS be a square of side length 
a + b as in Figure 9-15. The points T, (/, V, W are interior points of 
PQ, QR* R§, SP, respectively, such that FT = QU = RV = SW = a. 




T b Q 




380 Area and the Pythagorean Theorem 



Chapter 9 



Then TQ = UR = VS = W? = h, and ATQU. AURV, AVSW, 
A WFT are four congruent right triangles, each with area ^ih. Hence 
TUVW is an equilateral quadrilateral. Since die two acute angles in a 
right triangle arc complementary angles, it follows that 

nuLWTU =180- nUPTW - nUUTQ 

- mlPTW-mlPWT 

- (mlPTW+nUFWT) 
z 180 - 90 = 90. 

By a similar argument it can be shown that the other three angles of 
TUVW toe right angles. Therefore TUVW is a square. Let x be its side 
length. Then 

(a + fr) 2 = \PQRS 
(a + fc) 2 = 4 AWPTj + \TUVW\ 
{a + bf = 4 * \ab + x 2 
(a + &) 2 = 2al> + x 2 
a 2 + 2d/> + fc 2 = 2ab + x 2 
a 2 + fr 2 = x 2 . 

It follows from the S.A.S. Congmence Postulate that A WPT = 
AABG Therefore 

x = c and a 2 + fe* = e 2 . 

THEOREM 9.5 (Convene of The Pythagorean Theorem) If a, 
b, c are the lengths of the sides of a triangle and if a 2 + b 2 = c 2 , 
then the triangle is a right triangle and the right angle is opposite 
the side of length c. 

Proof: Let there be given AARC with AB - c, BC = o, CA = b, 
and a 2 + fc* = c; 2 . Let AA'B'C l>e a right triangle with B'C = a, 
CA' = b, A'B' = x, and the right angle at C" as shown in Figure 9-16. 




C o 

Figure SUA 




9.5 Pythagorean Theorem 381 

Then 

a2 + fez _ *2. Why? 

By hypothesis, a 2 4- h 2 = c 2 > By substitution, x 2 = c 2 and, since x > 
and c > 0, it follows that x-c. Therefore A ABC s AA'B'C'by the 
S.S.S. Postulate and LC^LC It follows that ZC is a right angle 
and that A ABC is a right triangle. 

When solving problems by use of the Pythagorean Theorem, it 
is often necessary to find the square mot of a number that is not a per- 
fect square. Recall that a rational number is a perfect square if it can 
be expressed as the square of a rational number. Thus 16 and 2 ,f are 
perfect squares because 

16 = (4)2 and V = ($)*, 

whereas 32 and -^ arc not perfect squares. If we arc required to find 
the square root of a number like 32 or -^- t we can approximate the 
square root by means of a rational number or we can leave our answer 
in what is called simplest radical form, If % = y/A, where A is a posi- 
tive rational number which is not a perfect square, we say that x is in 
simplest radical form when it is expressed as B>/C t where B is a ra- 
tional number and C is a positive integer which contains no factor 
(other than 1) that is a perfect square. 

We make use of the following theorems from algebra when putting 
radicals in simplest form, 

1. If a and b are positive numbers, then 

■\/ab — Vff * \/5* 

2. If a and b are positive numbers, then 

ft£_ Va 
V b yE' 

Example I Put (1) y/3£, (2) >/¥"' and (3) \/105 hi simplest radical 
form. 



i 



Solution: 1. V32 = \/W- 2 = y/TB • y/2 m 4 \/l" 

" / 8 " JT' 2 = X /I6 " 4 " 

3. \/I06 is already in simplest radical form. Why: 



382 Area and the Pythagorean Theortm 



Chapter 9 



EXERCISES 9.5 

In Exercises 1-15, put the indicated radical in simplest radical form. 

1- V% 9, \/2M 

& \/!8 10. \/L35 

3. v§7 

4. y^O 

5. v^S 

6. yfffi 

7. y300 



11. vf 
13. v*? 



x/ 12 



~rr 



16. If ci 2 + b 2 = r; 2 , solve for a in terms of fc and c, 

17. If o 2 + LP = c 2 , solve for b in terms of a andc. 
IS. If « 2 + b 2 = e 2 , solve for r; in terms of a and &, 

19. If the length of a diagonal ttf a square is 15, find the area of the square. 
{Hint: Use the Pythagorean Theorem,} 

20. In the figure, ABCD is a parallelogram, BE ± AB at E t AD - 15, 
CD = 20, and A£ - 9, Find the area olABCD. 




<> !■: 



20 



In Exercises 21-28, A ABC is a right triangle with right angle at C, a = BC, 
b = CA, c = AB, In each exercise, two of the three numbers a, b r c are 
given. Find the third one. Express each answer in decimal form correct to 
two decimal places. 

21. a = 14 h = 1 jO 
22- a = 1.0, b = 2.0 

23. a= 10.0 f fo= 10.0 

24. a = 3.0, 6 = 4.0 

25. a = 9,0, b = 40.0 

26. c = 3.1, fo = 5.2 

27. a = 10.0, fc = 9.0 

28. c ^ 10.0, a = 7,0 



9.5 Pythagorean Theorem 383 

In iLxereises 29-36, A ABC is a right triangle with right angle at C, a = BC t 

b = CA, G ss AM, In each exercise, two of the three numbers a, b, c are 
given. Find the third one, Kxpress each answer exactly and in simplest form, 
using a radical if needed. 

29. a - 7, b = 40 

30. a = 39, c = m 

31. a = \/2, h = vT 

32. = v/5I. fe = \ ^ 

33. a = J, fc = i 

34. o = 3x, £? = 4x (Find c in terms of *,) 

35. a = x + gft & = x — y (Find c in terms of x and y.) 

3ft. a = u 2 + C^ f a = a 3 — u 2 (Find b in terms of u and u.) 

37. See the proof of Theorem 9.4, From which Triangle Congruence Pos- 
tulate docs the conclusion that ATQV s A URV follow? 

38. See the proof of Theorem 9.4. From which Area Postulate does it fol- 
low that the area of the large square is the sum of the areas of the four 

triangles and the smaller square? 

39. Sec the proof of Theorem 9.4. From which Area Postulate does it fol- 
low that the four triangles have equal areas? 

40. In the proof of Theorem 9.4, which Area Postulate supports the con- 
clusion that the area of TUVW is x 2 ? 

41. Let AABC with BC = a, AC == b, AB = v, c>b,c>a, and 

c 1 > « 2 4" b 2 be given. Answer the "Whys?" in the following proof 
that L C is an obtuse angle- 




Proof; Let AA'B'C He a right triangle with B'C = a, CA' = b, 
A'B' = x, and the right angle at C as shown in the figure. Then 

a* + t>i s s* 

for AA'B'C, Why? By hypothesis, a 2 + ^ < # for A ABC. By sub- 
stitution, x 2 < c 2 and, since x > and o > 0, it follows that x < c, 
Therefore IC> LC Why? Since mZC' = 90, it follows that 
m£C> 90. Theu L C is an obtuse angle. Why? 



304 Area and the Pythagorean Theorem 



Chapter 9 



42. Let A ABC Willi BC - a t AC = b, AB = c, c > b, c > a, and c* < 
a 2 + fe 2 be given- Prove that / G is an acute angle. (See Kxercise 41) 

43. Combine Theorem 9.5 and the statement* proved in Exercises 41 and 
42 into a single theorem. 

44. chaijlenge problem. The Pythagorean Theorem may be stated as 

follows. 

In a right triangle the area of the square on the hypotenuse 
is equal to the sum of the areas of the squares on the legs, 

iiestatement! A ABC is a right triangle with the right angle atC, 
BC = a, AC = b, AB a c, squares ABDE, BCLK, and AIGC as 
shown in Figure 9- 17a. Prove Uiat 

t ABDE\ = \BCLK\ + \AFCX:\ 

and hence that c a = a 2 + h*. Complete the following proof which up» 
peared in Euclid's Elements around 300 b.cj. 





Figure ft.17 



Proof: Draw AK, CD, and €37 1 IM at M and intersecting A~B at X 
as shown in Figure 9-1 7b. 



Why? 



(1) AABK ^ ADBC 

(2) \AABK\ = \ADBC\ 

(3) The altitude from A to £3 of AABK is equal to BC. 

(4) AABK = | * KB - »C = } • fl 2 Whv? 

(5) JBCLiq =2*| AABK | Wh^ff 

(6) The attitLide from C to B3 of ADBC is equal to BN. 



(7) 'ACBD, = l-PB-JJA' 



Whv? 



Why? 
Why? 



Chapter Summary 385 

(8) BDMN\ = 2 * | ADBC\ Why? 

(9) \BCLK[ = \BDMN\ Why? 

(10) Draw BFand UE, ABAF = AEAC Why? 

(11) \AFGQ =2*|ABAf| Why? 

(12) |A£»fiV| = 2 • | A£Aq Why? 

(13) fAFCCf = [A£.\/,Y| Why? 

(14) ABDE = BDMN\ + \AEX1X Why? 

= \BCLK\ + AfGC| Why? 
(J 5) Therefore c 2 = « 2 + b 2 Why? 

45- challexce problem. 11 your high school library contains the book 
27*e Pythagorean Proposition, look up the proof devised by Miss Aim 
Gondii: and present it to your class. 



CHAPTER SUMMARY 

In this chapter there a re four Area Postulates and several area formulas 
stated ami proved as theorems. The last two theorems of the chapter are 
the Pythagorean Theorem and its converse. 

POSTULATES 

Area Existence 
Rectangle Area 
Area Congruence 
Area Addition 

FORMULAS 

Rectangle Area Formula S = ah 

Parallelogram Area Formula S s hh 

Triangle Area Formula S = \bh 

Trapezoid Area Formula S = -J(&i + b 2 )h 

THE PYTHAGOREAN THEOREM. If a, b, c are the lengths of the 
sides of a right triangle AA/JC, with c = AB> the length of the hypot- 
enuse > then a 2 + b 2 = c 8 . 

You should be able to stale each of the postulates in your own words 
and to explain the meaning of each of the formulas. You should be able to 
use the postulates in developing the properties of area. You should under- 
stand the formulas so that yon can recognize a situation in which a formula 
is applicable and then use it correctly. 



386 Area and the Pythagorean Theorem Chapter 9 

REVIEW EXERCISES 

1. Find the area of a trapezoid with two parallel sides of lengths H and 13 
if the distance between the lines containing those sides i- 24. 

2. If the hypotenuse of a right triangle is 50 in. long and one of the lcg?i is 
40 im long, how long is the other leg? 

3. The figure shows a right triangle A.ABC with right angle at A and with 

D the foot of the perpendicular from A to EC. If BD - 3, DC = 5{, 
and Ai3 = 5, find \ A ABC 




4. Find the area of a right triangle if the length of the hypotenuse is \/T 
and the length of one log is y/5. 

5. Find the base of a parallelogram if its area is 143 and its height is 7. 

6. Find the altitude of a triangle corresponding to a base of length 12 if 
the area of the triangle is 62, 

7. If A ABC is an equilateral triangle of side length 6. find the area of the 
triangle. (f/iitfcUsc the Pythagorean Theorem to find an altitude of the 
triangle.) 

8. See Exercise 7. Prove that the area S of an equilateral triangle of side 
length f is given fay 

■ -J&.A 

4 

9. Use the result of Kxercise 8 to find the area of an equilateral triangle 
of side length 12, Compare the area of this triangle with that of Kxercise 
7. Is the ratio of the areas of the two triangles the same as the ratio of 
the lengths of their sides? If not, how do the areas compare? 

10. If ABCD is a parallelogram with m/ DAB — 45, DA a 12, and 
DC = 21, find \ASCD\. (Mint: Draw ~DE _ AB at E, Which kind of 
triangle is A DAE?) 

In Exercises 11-13, A ABC h given with a = EC, h = AC. and c = Afl 
In each exercise the three numbers a, b, c are given. Is L C a right, an ob- 
tuse, or an acute angle? (See Exercise 43 of Exercises 9.5.) 

11. a = ft6 = g,cs 12 

12. a = 8, b = 15, c = 17 

13. a = ll,b= 13, c = 17 



Ftevtow Exercises 387 

14 *!Tie area of a trapezoid is 164 and the distance between the parallel 
bases is 10. If the length of one of the bases is 15, find the length of the 
other base. 

15. (An informal geometry exercise.) The dimensions of a shower stall 
are 3 ft. by 3 ft. by 7 ft. How many 4 in. by 4 in. tiles are needed to 
cover throe of the four rectangular walls if wc assume that there is no 
waste in cutting the tile? 

16. {An inform al geometry exercise, ) \ rectangt. Jar plot of land is 1 00 yd. 
by 150 yd. A standard city lot for this particular plot of land is 75 ft. 
by 150 ft. and sells for $3000- If a real estate agent can buy the plot of 
land for $30,000, how much profit would he make if he divided it into 
standard lots and sold all of thein? Disregard surveying and legal 
expenses, 

17. An isosceles triangle has two sides of length 13. If the altitude to the 
base is 12, find the area of the triangle. 

18. A rhombus has sides of length 10 and the measure of one of its angles 
is 45. Find its area. 

19. The length of the hypotenuse of a light triangle is 25 and the length of 
one leg is 24, 

(a) Find the length of the other leg. 

(b) Find the area of the triangle. 

(c) Find the altitude to the hypoteuusc. 

20. A right triangle has legs of lengths 10 and 24 and hypotenuse of length 

(a) Find the area of the triangle. 

(b) Find the altitude to the hypotenuse. 

21. The length of one base of a trapezoid is .5 more than twice the length of 
the other base. If the area is 100 a net the altitude is 10, find the lengths 
of the two bases. 

22. In the figure, AD is an altitude of A ABC. If AD = 24, AB a 26, and 
AC = 30, find BC and A ABC. 




23. Recall that the diagonals of a rhombus are perpendicular to each other. 
Use this fact to find the area of a rhombus whose diagonals are of lengths 
10 and 8, How is the area of the rhombus related to the lengths of its 
diagonals? State this relationship in the form of a theorem and prove it. 



388 Area and the Pythagorean Theorem 

24. In the figure, All _l W t B-C-D, AC = 10, BC 
FindHDand jAAG'D|, 



Chapter 9 
AD = 17. 




25. In the figure, with right angles and congruent segments as marked, if 
VA = 2andAB = 1, find VF. 




26. Taking the area of a given unit square as I and using any of the Area 
Postulates except the Rectangle Area Postulate, prove that the area of 
a 4 by 5 rectangle is 20. 

27. Taking the area of a given unit square as 1 and using any of the Area 
Postulates except the Rectangle Area Postulate, prove that the area of 
a i by i rectangle te £■ 

28. The figure suggests an alternate proof of the Pythagorean Theorem 
which makes good use of the area formulas for a trapezoid and a tri- 
angle, Write out the main steps in this proof. (The proof suggested by 
this figure is attributed to President Garfield and was discovered by 
him around 1876.) 




Review Exercises 



381 



29. challenge problem. The figure shows a triangle and its three medi- 
ans intersecting at the point G which is two-thirds of the way from any 
vertex to the midpoint of the opposite side. Prove that 




30. challenge problem. Given the same situation as in Exercise 29, 

prove that the six small triangles all have the same area. 




Courtesy uf Leo GatieJU Gallery 



Similarity 



10,1 INTRODUCTION 

Our experiences with objects of the same size and shape suggest 
the concept of congruence in formal geometry. The idea of same shape 
suggests the concept of similarity which you are ftbouA to study in this 
chapter. Consider a picture and an enlargement of it. In the enlarge- 
ment each part has the same shape as it has in the original picture, but 
not the same size. The picture and its enlargement are examples oi 
similar figures. 

Consider a floor plan for a building. The floor plan is a drawing 
made up of segments labeled to show the lengths of the segments in the 
actual building which the segments on the drawing represent. The 
floor plan is similar to the actual floor of the building. Although the 
plan and the floor do not have the same size, they surely have the same 
shape. Each segment in this plan is much smaller than the segment it 
represents in the building. We express this fact by saving that the plan 
is a scale drawing of the floor. Each angle in the plan has the same size 
as the angle it represents on the actual floor. 

In drawing a floor plan, lengths arc reduced and angle measures are 
preserved, The lengths of segments in a floor plan are proportional to 
the lengths of the segments that they represent in the floor, and a state- 
ment of this fact is an example of a proportionality. In the next two 
sections we define a proportionality and develop some of its properties. 



392 Similarity 



Chapter 10 



In the remaining sections we develop the concept of similar figures. 
The main theorems of this chapter have to do with triangle similarity. 
The chapter includes a proof of the Pythagorean Theorem based on 

similar triangles. 



10.2 PROPORTIONALITY 

Figure 10-1 shows two figures, one with side lengths labeled a, b, 
c, d, e and the other with a\ b\ c\ d\ e\ Note that a = 2a\ b = W t 
c = 2c\d = 2d", e = 2e\ and hence that 

JL — JL — .£ , d__ e_ __ p 

a' "" V " c' " d' " e f " 



dmH 



c-3 



•- ?■ 9 




c'*L5 



d'mS 



*' = as 



a* 4 




a'm2 



e'«4 



Figure 10-1 



We can express the idea of these equations by saying that a, b, c, d r e 
are proportional to a\ b\ c\ d\ e\ respectively. 



Definition J 0.1 Let a one-to-one correspondence between 
the real numbers a, b, c, . . . and the real numbers a',h\ c f t 
... in which a is matched with a', h is matched with b\ c is 
matched with c\ and so on, be given. Then the numbers a, h, 
c, ... are said to be proportional to the numbers a', b\ c', . . . 
if there is a nonzero number k such that a = ka\ b = kh\ 

c = k& The number k is called the constant of 

proportionality, 



in Definition 10.1, why do you think k is required to be a nonzero 
number? 



10.2 Proportionality 393 

Notation. We use = to mean "are proportional to." We use 
(a, b> c s . . .) to represent an ordered set of numbers. Then 

w 
means that the numbers a,b,c,... are proportional to the numbers 

d % h\d, t it being understood that a is matched with a\ h with f, 

c with c', and so on. Note that the order of the numbers in the set 
{a, h, c, , , ,) and in the set [a', b', c', . . .) is important only to the extent 
that corresponding numbers must appear in the same order in the pro- 
portionality. Thus if we have 

we could also write (h, c, a, . . .) = (bf, <f $ a\ . . .), and so on. 

Example 1 (See Figure 10-1.) (4, 7, 3, 6, 8) = (2, 3.5, 1.5, 3, 4). 

Example 2 (2, 3.5, 1.5, 3, 4) = (4, 7, 3, 6, 8). 

Since 

2 = ^-4, 3.5 = 4-7, 1-5 = 4*3, 3 = ^-6, 4 = ^-8, 

the constant of proportionality in Example 2 is j. What is the constant 
of proportionality in Example 1? 

Example 3 If (*, y t 9) = (7, 8, 6), find x and y. 

Solution: There is a number k such that % = k ■ 7, y = k • 8, 
9 = k • 6. From the last equation we have k = j, so 

3 = 4-7 = 10.5 and y = §-8=12. 

Example 4 U the following statement true or false? 

(5, 8, 10) = (8, 12.8, .16). 

Solution; The statement is true if and only if there is a nonzero num- 
ber k such that 

5=Jfc-8 

8 = it- 12.8 
10 m k* 16 

From the first of these three equations, we see that k must l>c -|, or 
0.625. We must now check to see if this k "works" in the other two 
equations, 



394 Similarity Chapter 10 

Check. Is 6 = 0,625 - 12.8? Yes, 
Is 10 = | • 16? Yes. 
Therefore (5, 8, 10) = (8. 12.8, 16) is true. 

Example 5 Ts the statement (4, 8, 10) = (6, 9, 16) true? 

Solution: 

If 4 = k • 6, then Jfc = f , 
Is 6 = j< 9? Yes, 
Is 10= | -16? No. 

Is the statement (4, 8, 10) = (6, 9, 16) true? No. 



EXERCISES 10.2 

In Exercises 1-12, determine whether the given statement is true or false. 
If it is true, find the constant of proportionality. The answer for Exercise I 
has been given as a sample. 

1. (1, 1, 5) = (3, 3, 15). TYue; k = \. 
E. (3, 6, 9) = (2, 4, 6) 

3. (5, 1) = (8. 2) 

4. (8, 10, 15, 21) = (7, 1 1. 16, 22) 

5. (5, 15, 25, 35) = (1, 3, 5, 7) 

6. (2, 6, 7, 15) = (1, 3, 3.5, 7.5) 

7. (1,3.1 +3) = {2, 6, 2 + 6) 

8. (x, y. % + y) = (7x, ly t 7x + 7y) 

9. (m, v,w) = {-3«, -3t\ -3m>) 
10, (0, 0, 0) = (5, 10, 15) 

U. (5, 10, 15) = (0, 0, 0) 

12. (0, 0, 3) = {0, 0, 4) 

In Exercises 1 3-20, copy and complete the given statement so that it will 
be true. 

13. (8, 8) = (0,6) 

14- mm, 100) = (5, 6, 10) 

15. (5, 0,7) = ([7], 8, 10.5) 

16. If (3, x) = (8, 10), then x = [7J. 

17. If (7. 9) = (1x, 18), then a = [7J. 

18. If (7. x) = (14, 21), then x = \t\> 

19. If (5 t i) = {*, 125) and * > 0, then x = [7]. 

20. If (3, x) = (8, if) and i/ yt o t then {3, [3 ) = (*, y). 



10.2 Proportionality 395 

In Exercises 21 -30, x is a positive number and (3. x) = (5, 15). In each ex- 
ercise, determine whether the given statement is true. 

21. £ = ^- 

5 15 

22. — = — 
x 15 

23. (3,*.3 + *) = {5, 15*20) 

24. {x, 3 - x) = (15. 10) 

25. (*, 3) f (15, 5) 

20. (3, 5) = (x, 15) 

27. (3. 15) = fc 5) 

28. (3 t 5) = {15\r) 

29. 3-15 = x-5 

30. 



3 5 



x + 3 5+15 

31. Does (fl, &, c) * — » (the,/) indicate the same one-to-one corre- 
spondence as (a, c, b) * — * (d, /, *?)? As (c, a, fo) « — * (/, rf T e)? As 
fcfea) *— * (/.«.d)? 

32. If a, ft, e, <i are nonzero numbers such that (a, h) = [c, d) with pro- 
portionality constant 2 , is it true that (b,a) = {d, c) with proportionality 
constant 2? Give your reasoning. 

33. If a, b, c,d are nonzero numbers such that (a, b) = {c, d) with propor- 
tionality constant 2, is it true that (c, d) = (a f b) with proportionality 
constant 2? Give your reasoning. 

34. If a t b, c, d are nonzero numbers such that (a, b) = (c, </), prove that 
(a, c) = {b, d). 

35. If {a, b) = (c, if), prove that ad = he, 

36. If ad = fee, prove that {a, b) = (c, d). 

In Exercises 37-40, copy and complete I he given statement so that it will 
be a proportionality. 

37. (5.7,10.12) = (10, [Sm, 0) 

38. (0, H, 1) f (15* 37, 100) 

39. (0,37. 0.67, 0.93) = ((?], 67, fTJ) 

40. (V5 v^V^f CV&EH3) 



396 Similarity Chapter 10 

10.3 PROPERTIES OF PROPORTIONALITIES 

As you might expect from your study of algebra and from some of 
Exercises 10,2, proportionalities have some interesting properties. In 
this section we show that the proportionality relation is reflexive, sym- 
metric, and transitive, and therefore it is an equivalence relation. 
Hence the relation denoted by **=" has some properties in common 

r 

with the relations denoted by " = " and "^." We shall also state and 
prove addition and multiplication properties. At the end of this sec- 
tion we consider some special proportionalites called proportions. 

We shall prove the next two theorems for proportionalities involv- 
ing triples of numbers. It is easy to see how the statements and proofs 
can be modified for proportionalities involving more than or fewer 
than three numbers. 

THEOREM 10.1 The proportionality relation is an equivalence 

relation. 



Proof: 



1. The Reflexive Property. Let a, b* c be any real numbers. Ob- 
serve that (a, b f c) == (a, b, c) with proportionality constant 1. 

2. The Symmetric Property. Suppose that {a, h % c) — (*£» e,/). Then 
diere is a nonzero number k such that a = kd, b = ke t c = kf. 

Why? Then d = k'a, e = k'b, f= k'c, where k' m l and 

k' ffe 0. Therefore (d, <?,/) = (a, b, c). Thus if (a, b, c) = (d t ej) 
with constant of proportionality k t then (d. e t f) = (a, b, c) with 

constant of proportionality — . 

3. The Transitive Property. Suppose (a, b* c) = (d t e t /) and 
{d, e f f) = (g, h y i). Then there are nonzero numbers kj and k 2 
such that 

a = kid, b = k\e, c = k\f, 
d = k 2 g, § = k 2 K f = M- 



Then 



a = kid = fci(feg) as (hk 2 )& 
b = k x e = ki(k 2 h) = (hk 2 )h, 
c= kif=ki(k 2 i) = (k t k 2 )i. 



10.3 Properties of Proportionalities 397 

Since In jfi and k z # 0, it follows that Mfc f£ 0, Therefore, if 
a, b t c are proportional to d, e, / with constant of proportional- 
ity &i, and d, e./are proportional to g, ft, i with constant of pro- 
portionality fca, then a, b, c arc proportional to g, h, i with con- 
stant of proportionality fci& 2 - 

In the next theorem note that (1) is an "addition" property and that 
(2) is a "multiplication"' property. 



THEORRV 10.2 If (a, b, c) =: (</, e f /), then 

(1) ( 
and, if h ^ 0, 



(1) (a, fe, c. a + b + c) = {d t e t f, d + e + f) y 



(2) {ha t b>c) = {hd>e t f). 

Proof: Let it be given that (a> b, c) = (ef t e, /). Then there is a non- 
zero number k such that 

a = fcd, b = fee, c = Jfcf. 

Adding, we get 

a + b + c e faf + he + fcf, 

and, by the Distributive Property, 

a + b -*- c = k(d + e + /). 

Multiplying both sides of a = kd by h, we get ha = h(M), and hence 

ha =s Jt(fcd). 

We have shown that if a. b % c are proportional to d, e, f with constant 
of proportionality k, then (1) and (2) hold with the same constant of 
proportionality. 

Example 1 Figure 10-2 shows a triangle A ABC and a segment DE 
joining a point D of AB to a point £ of AC. Suppose we know that 
{AD, DB) = (A£, EC). 



Figure 10-4 




398 Similarity Chapter 10 

We may conclude from the addition property of proportionality that 

{AD, OB, AD + DB) = (AS, EC, AE + EC) 

and hence that 

(AD, DB, AB) = (AE, EC, AC), 

since AB = AD + DB and AC = A£ -f- EC. Of course, we may also 
conclude that 

(AD, AB) j AE, AC). 

This may not be the "whole truth/' but it is certainly the "truth." It is 
like concluding that if x = 3», y — 3u, and z = Bm, then * = Bit and 
y = Sv. Similarly, we may conclude that 

(DB, AB) = (EC, AC), 

Example 2 Let A ABC with D an interior point of BC be given as 

suggested in Figure 10-3. Let /* denote the distance from A to BC, let 
BD = bu and let DC = b* Then 



| AABD\ = * 2 hbu \AADC\ - $hb 2 . 




Figure 104 B 



Therefore (\AABD\, \AADC\) = (bi, bz), and the constant of propor- 
tionality is -j/i. Thus the areas of the two triangles formed from A ABC 
by inserting the segment AD are proportional to the lengths of the seg- 
ments formed from EC by inserting the point D, In this connection 
some of you will recall Challenge Problem 29 in the Review Exercises 
of Chapter 9, 

We frequently work in geometry with proportionalities in which 
two numbers are proportional to two numbers. Such proportionalities 
are generally called proportions. Following a formal definition is a 
list of some of their special properties. 



Definition 10.2 If a t b, c, d are numbers such that 
(a> b) =: (c, <l) is a proportionality, then that proportionality 
is a proportion. 



10.3 Properties of Proportionalities 399 

lii other words, a proportion is a proportionality with two num- 
bers on each side of the " — " symbol. 

The following theorem includes four named properties of 
proportions. 

THEOREM 10.3 Proportions involving nonzero numbers a, h, 
c, d have the following properties: 

1. Alternation Property. If (a, b) = (c, d), then 

(a, c) = (b t d) and (cf, h) = (c, a). 

2. Inversion Property. If (a , fr) =3 (e» d) % then 

(h a) = (</, o). 

3. Product Property, {a, b) = (c, d) if and only if ad = be. 

4 Ratio Property, (a, b) = {c, d) if and only if -y = ■£ . 

p b a 

Proof: Assigned as exercises. 

If you think of a and d as the "outside" numbers and b and c as 
the "inside" numbers of a proportion 

(*. *) f te A 

the Alternation Property says that if you interchange either the out- 
side numbers or the inside numbers in a proportion, then the result is a 
proportion. 

If you combine the Inversion Property with the Ratio Property, 
the Inversion Property amounts to saying that if two ratios are equal, 
then their "inversions" (reciprocals) are equal. 

In Theorem 10.3, the numbers in the proportions arc required to 
he nonzero numbers. What would lie the situation if zeros were per- 
mitted? Note that (0, 5) = {0, 10} is a true statement, whereas (0, 0) = 
(5 T 10) is a false statement. Therefore the proportion 

(0, 5) = (0, 10) 

does not have the Alternation Property. Observe that « 7 s • 8 is a 
true statement, whereas (0, 0) = (8, 7)isa false statement. Therefore 
the Product Property does not apply to • 7 = ■ 8. Observe that 

a,0) = (2,0) 

is a true statement, whereas ^ = -^ is a false statement. Therefore the 
Ratio Property does not apply to (1, 0) = (2, 0). 



400 Similarity Chapter 10 

EXERCISES 10.3 

1. If 0, b t c t d are all nonzero numbers, then (a t h t c t d) — (a, &* c, d). 
What is the constant of proportionality in this proportion? Which prop- 
erty of an equivalence relation does this illustrate? 

2* Given {a, b, c, d) = faj\ g> h), prove that 

fa f, g. h) = fa b> c d). 

Which property of an equi valence relation does this illustrate? What 
is the relation between the constants of proportionality for these two 
proportionalities? 

3. Given (a, b) = fa d) with proportionality constant k\ and (c, d) = faf) 
with proportionality constant k->, then fa b) = fa f) with some pro- 
portionality constant, sny k^. How are k±, hit, k-t related? Which prop-- 
erty of an equivalence relation do we have illustrated here? 

4. In the figure A, B, C are noncollinear points; A, B\, B& B are eollinear 
and arranged in the order named; A, C u C 2i C are colUneur and ar- 
ranged in the order named. It is given that 

(AB U »,B 3 . B 2 B) = (AC,, C,C 2 , dQ 

with constant of proportionality 0.8. Prove, using the properties of 
proportionalities, that 

(a) (AB,,AB) = (AC lt AC). 

(b) (16AB U AB) = (I6AC1, AC), 




In Exercises 5-11, complete the given statement and name the property 
which it illustrates. Assume that none of the numbers in these exercises is 
zero. 

5. If fa b) = (3, 5), then (a, 3) = (£>, 0). 

6. lf(x,i/) = (6 3 7) t then ( ?/ ,,v) f (7,0). 

7. If (m. c) = (5, 6) and (5, 6) = (x, tj), then [u, v) = (x, [?]). 

8. If (a. b) = fa 00), then fa a + b) = fa c + d). 

9. If (*, y) = (I, 2X then (3*. y) = {[[J, 2). 

10. If (x, y) = (J7J, 7), then 7* = 4i/. 

11. If (*, y) = (5, 8), then * = [B 



10,4 Similarities between Polygons 401 

In Exercises 12-21, write a proportion, or complete the given one, so that 
it will be equivalent to the given information. Starting with the given in- 
formation, you should be able to prove that the proportion is true. Starting 
with the proportion, you should be able to prove that the given equation 
(or equations) is true. 

16. a = 3*. b = 3y 21. ^ = - 

Exercises 22-25 refer to Theorem 10,3. Prove that proportions involving 
nonzero numbers have the indicated properties. 

2SL The Alternation Property. 

23. The Inversion Property. 

24. The Product Property. 

25. The Ratio Property. 



10.4 SIMILARITIES BETWEEN POLYGONS 



In this section we define what is meant by a similarity between two 
polygons. In developing our formal geometry we use similarities as 
tools; in most cases we consider similarities between triangles. 

Definition 10.3 

1. A one-to-one correspondence ABC . , . * — > A'B'C , . . 
between the vertices of polygon ABC . . . and polygon 
A'B'C ... is a similarity between the polygons if and 
only if corresponding angles are congruent and lengths of 
corresponding sides are proportional. 

2. If ABC . . . < — * A'B'C ... is a similarity, then polygon 
ABC . . . and polygon A'B'C , , . arc similar polygons and 
each is similar to the other, 

3. If ABC . . . a — * A'B'C ... is a similarity with 
A3 = kA'B', BC = kB'C, and so on, then k is the con- 
stant of proportionality, or the proportionality constant, 
for diat similarity. 



402 Similarity 



Chapter 10 



Notation, The symbol "W is read "is similar to"; hence ABC ,..*-» 
A'B'C ... is read "ABC . , . is similar to A'B'C . ..." This means that 



A IH 



A'B'C 



Example 1 Fi give 10-4 shows two quadrilaterals ABCD and A 'B' CW 
with segment lengths and angle measures as indicated. It ap- 
pears that ABCD * — * A'B'CD' is a similarity and hence that 
ABCD ~ A'B'CD'. Let us check this conjecture. There arc eight pairs 
of corresponding parts including four pairs of corresponding angles 
and four pairs of corresponding sides: 



LA 


and I A'; Al5 


and AT*'; 


B 


and L W-. BE 


and JfC; 


LC 


and AC; UD 


and CT?j 


LD 


and L D*; UA 


and /XV. 


D A 


C 




DO 

1 


I34X 




90 
1 


X 


Ag/^ 


A 


7 B 


P 


FIgun KM 






First we check the angles. 






mZ.A = niLA' = 


90, 




mlB s mLB' = 


45, 




mLC = mLC = 


135, 




mLD = mLD' = 


90. 



Therefore, corresponding angles are congruent. Next we check to see 
if the lengths of corresponding sides are proportional. We want 

(AB, BC, CD. DA) = (A'B f . B'C, CD\ &A% 

Substituting, we get 

(7,3x/2*4, 3) = (3.5, 1.5 y/2, 2, 1.5). 

Since this is indeed a proportionality, with constant of proportion- 
ality 2, the lengths of corresponding sides arc proportional. There- 
fore, it follows dircctJy from the definition of a similarity that 
ABCD- A' B'C D\ 



10.4 Similarities between Polygons 403 

In Example 1 the constant of proportionality between the lengths 
of the sides of quadrilateral ABCD and the lengths of the correspond- 
ing sides of quadrilateral A'B'CD' is 2. Is it possible that in another 
example the constant of proportionality might be 1? Of course it is. In 
this special case of a similarity, corresponding angles are congruent 
and corresponding sides are congruent. Hence for this special case you 
should sec that Lhe similarity is a congruence. In other words, a 
congruence between polygons is a similarity between polygons for 
which lhe constant of proportionality is L 

Just as congruence for triangles is an equivalence relation , so also 
is similarity for polygons. We slate this fact as our next theorem, 

THEOREM 10 A The relation of similarity between polygons is 
reflexive, symmetric, and transitive. 

Proof: We shall prove the theorem for triangles. Tt is easy to modify 
this proof to get a proof for quadrilaterals* pentagons, and so on. 

Reflexive. Let AABC be given. Since AABC at AABC, it follows 
that A ABC — A ABC. What is the constant of proportionality for 

(AB, AC, BQ = (AB, AC, BQ? 

Symmetric. Suppose that AABC — ADEF; then ZA^ZA 
ZB^ZJE, ZC==ZF,and 

{AB, BC, CA) = (DE, EF, Dl<)> 

We want lo prove that ADEF — AABC, which means that ZD^ 

LA, LE^i LB, LF= / C, and that 

(DE, EF, DF) = (AB, BC, CA). 

Now LA=*LT) implies LD^LA, LBsLE implies LM m£B, 
and £Cm£W implies LF = Z C (Which property of congruence 

for angles supports tills deduction?) Also, 

(AB, BC, CA) = (DE, EF, DF) 

implies Hi at 

fVm, EF, DF) = (AB, BC, CA). 

(Which property of proportionality supports this deduction?) 
Therefore 

AABC — ADEF 
implies that 

AD£F- AABC, 

and similarity for triangles is a symmetric relation. 



404 Similarity Chapter 10 

Transitive* Let it be given that AABC ~* ADEF and that 
ADEF — &GHL The proof that A ABC ~ AGH1 is assigned as an 
exercise. From tills it follows that similarity for triangles is a transitive 
relation, 

THEOREM 10,5 The perimeters of two similar polygons arc 
proportional to the lengths of any two corresponding sides. 

Proof: As in the case for Theorem 10.4, we shall prove Theorem 10.5 
for triangles. It is easy to modify this proof to get a proof for quadri- 
laterals, pentagons, and so on, 

Ijet A ABC — AA'B'C be given and let p be the perimeter of 
A ABC and p* the perimeter of AA'B'C, By the definition of similarity 
we have 

(AB, BC, CA) = (A'B\ B'C, CA'). 

By Theorem 10,2, 

(AB, BC,AC, AB + BC+ CA) 

= {A'B\ B'C, A'C, A'B' + B'C + CA'). 

Bui 

p = AB + BC + CA and p* = A'B' + B'C + CA', 

Therefore 

(AB, BC, CA. p) = (A'B', B'C, CA\ p% 

It follows that 

(AB, p) = (AW, jf) 

and, from the Alternation and Inversion Properties that 

(p,p') = (AB.A'ff). 
In a similar way, it can be shown that 

(p, p') = (BC, B'C) and (p, p') f (CA, CA'). 
This completes the proof of Theorem 10.5 for triangles. 

EXERCISES 10.4 

1. Complete Ihe proof of Theorem 10.4 by proving thai 
A ABC- AGHl 

2. Given A ABC - ADEF, mLA = 30, m£B = 60, AB m 20, 
BC = 10, CA = 10 V3, DE = 100, find mlD, mlE, mZF, EF, 
and DF, 



10,4 Similarities between Polygons 405 

In Exercises 3-10, ihcre are figures showing two similar triangles with some 
side lengths indicated and a similarity statement given. In each case, find the 
proportionality constant of the given similarity statement and Uks lengths 
of any sides whose lengths are not given. 

3. 





A ABC- 6D£P 



4, A 





a P 7 S S q 



A ADC - AACB 




&PQR- ATQS 



7, T 




ATSZ - AXV7 



406 Similarity 
8. D 



Chapter 10 




9. J? 




t l A 



ARKD- A HAT 




&ABC - AA'R-'C 



11. Given: A ABC <~ AA'B'C 

AA'B'C — AA"£"C" 
AB = JA'S* 
A'W - 5A"B" 
Pmce: AABC =g AA"J?"C" 

12. Let AAXT — AAtt/D wilh 



A\ AT AT 
HU AfD DD 



= 1000. 



If the shortest side of AMTflD is of length 1000, find the length of the 
shortest side of A A AT. 



10,4 Similarities between Polygons 407 

13. II in Exercise IS, p and p' denote the pcriinclcrs of the smaller triangle 

and tile larger triangle, respectively, find *— ■ 

P 

14. ABCDE and A'B'UD'E' are pentagons such that 

(AB t BC, CD, DE, EA) = (A'B\ WC t CD f t &E% E'A% 

£A=*£A\ LBzt£W t £Cm£O a 
LD^LD\ £Bat£&, and AB = ISA'S 1 . 

Prove that the perimeter of ABCDE h 13 times the perimeter of 
A'B'CD'E'. 

15. If A ABC ~ AA'B'C and AB = 10, HC = 8, A'fl' = 25, A'C = 35, 
find B'C and AC. 

16. If AFQR ~ ASTV and PQ = 24, ST = 16, PR = 18, TV = 10. find 
Qfl and SV. 

17. If A ABC ~ ARST and 7 ■ AB = 4 « RS, what is the ratio of the per- 
imeter of A ABC to the perimeter of ARST? 

18. If A ABC- ADEF, AB = 5, BC = 7, AC = 8, and the perimeter 
of ADEF is 60, find DE, £F t and DF. 

19. In the figure, A-K-T-S, RK ± S3, RK = 3. AT = 8, 15 = 4. 




(a) Find | AHAT|, the area of A RAT. 

(b) Find |AflTS|, 

(c) Solve for at: (| A RATI, |ARIS|) = (,v, 4) 

20. In the figure, A-K-T-S> M. 1 A3, JW: = 3, AT = 12, TS = 2. 




(a) Find [ARAT|. 

(b) Find ,AHTS|. 

(©) Solve for *: (*, ARTS|) = (12, 2). 



408 Similarity Chapter 10 

21. In the figure. Af-S-O-T, M J. S7T, MO = 12, OT = 5. 




(a) Solve for xs flAiVAfq, A.VOTJ) = (12, *). 

(b) Solve for tjx (\ANMO\. y) = (|A.VOT| P 5). 

In the figure, R-F-S-T, R-D-E-W, SE || TW t FE ± KT, D5 ± EW, 
RS = 20, ST = 10, RE = 30, EW = 15, FE = 29. 




(a) Find \ARSE\, (b) Find | AST£|. 
(c) Find SD, (d) Find | A SEW]. 

In the Bgure, A-D-B, A-E-C, Z5E || BC, £D = 12, DA 
\&BDE\ = 120. 

(a) Find;AADE|. 

(b) FirtdjADECj. 

(c) Find the ratio of AE to EC. 

(d) Find the ratio of AD to DB. 

(e) Compare the ratios in (c) and (d). 




10,5 Some Length Proportionalities 409 



10.5 SOME LENGTH PROPORTIONALITIES 

In this section there arc several theorems regarding the lengths of 
segments formed by lines parallel to one side of a triangle intersecting 
the other two sides, and there are other theorems extending these ideas 
to lengths of segments formed when three or more parallel lines are 
cut by two transversals. These theorems are useful in proving the 
similarity theorems of Section I0;6. 

THEOREM 10.6 (Triangle Proportionality Theorem) If a line 

parallel to one side of a triangle intersects a second side in an in- 
terior point dien it intersects the third side in an interior point, and 
the lengths of the segments formed on the second side are propor- 
tional to the lengths of the segments formed on the third side. 

Proof: Let there be given a triangle A ABC and a line / such that I 
intersects KB in an interior point D and such that t is parallel to EC as 
suggested in Figure 10-5. We know from Theorem 2.6 that Winter- 
sects A(7 in an interior point Call it £. We shall prove that 

(J5D, AD, BA) = (CE, A£, CA). 

A 




Figure 10-5 



Ut 



hi be the distance from E to AB, 
/*2 be the distance from D to AC t 
ft 3 be die distance between t>E and BC. 

Following are the main steps in the deductive reasoning which com- 
pletes the proof of the theorem. You are asked in the Exercises to sup- 
ply the reasons for steps 1 through 7 (definitions, theorems, preceding 
steps; etc.). 



1. 



BD ^hj-BD _ ABED\ 
AD ~ ^i t - AD " \AAED\ 

_ CE Mio-CE _ \ACED\ 
* AE " %hg'AM ' AAED 
3. | ABED\ = ^ 3 • DE = \ACED\ 



410 Simitar ity Chapter 10 

4 BD = CE 
' AD AE 

5. BD-AE = AD'CE 

6. (BD, AD) = (CE t AE ) 

7. {BD, AD, BA) = (CE, AE, CA) 

Therefore, the lengths of KB and the segments formed by I cutting 
KB are proportional to the lengths of 7l€ and the segments formed 
by I cutting JlC. From step 7 we get step 8, 

8. {BD, BA) s (CE, CA) and (AD, BA) = (AE, CA) 
From step 8 we got step 9 using the Alternation Property. 

9. (BD, CE ) = (BA, CA) and (AD, AE) = [AB, AC) 

Therefore, the lengths of AB and the segments funned by I cutting 
other two sides in interior points, then it exits oif segments whose 
lengths are proportional to the lengths of those sides. 

THEOREM 10.7 (Convene of the Triangle Proportionality 
Theorem) Let A ABC with points O and E such that A-D-B and 
A-E-C be given. If 



(AD t AB)~(AE t AC), 



then DE I BC. 



Proof: Let AABC with points D and £ such that A-D-B, A-E-C, 
(AD t AB) =; (AE t AC), as suggested in Figure 1Q-6, be given. 




B 

Figure MMSI 

Let I be the unique line through D and parallel to BC. Let W be the 
point in which ] intersects AT. In the figure, E and A" appear to be dif- 
ferent points. We shall show that they are actually the same point. 

Since I is parallel to BC, It follows from Theorem 10-8 that 
(AD,AB) = (AE\AC). 



10,5 Some Length Proportionalities 411 



But it is given that 

Therefore 
Then 



(ADtAB) == (AE.AC). 



(AE\AC) = (A£, AC). 



Why? 



AE'-AC = AE-AC (Why?) 

and AE' = A£. Since E and £' arc points of AC that are on the same 
skle of A and at the same distance from A* it follows that £ = E'. Since 

/ = ££' = M t 

it follows that BE || 5C. 

The word "cut" is used frequently as a synonym for "intersect/' 
particularly in situations involving several lines and a transversal,, as in 
our next theorem. 

THEOREM 10.8 If two distinct transversals cut three or more 
distinct lines that are coplanar and parallel, then the lengths of the 
segments formed on one transversal are proportional to the lengths 
of the segments formed on the other transversal. 

Proof: We shall prove the theorem for three p. ir.il lei lines. It is easy 
to modify the proof for more than three parallel lines. 




V 



Figure 10-7 

Let * and / be two distinct transversals of three distinct lines (, m, 
n that are coplanar and parallel as suggested in Figure 10-7. Label the 
points of intersection as in the figure. Draw XF. Let G be the point 
of intersection of m with AF. It follows from the Triangle Proportion- 

:tlt(\ Tli or .rem. applied to ;\ACh, rhat 



I: 



{AB > AC) = (AG t AF) 



and from the same theorem, applied to &AFD, that 



(2) 



(GA, FA) = (ED, FD). 



412 Similarity Chapter 10 

From the Transitive Property of Proportionality it follows that 
:V3) (AB t AC) = (DE,DF). 

Similarly, we may show that 

(4) (BC t AC) = (EF,DF). 
Then from (3) and (4) we gel 

(5) (AB, BC, AC ) = (DB, EF t DF ). 

COROLLARY 1Q.8J If a line bisects one side of a triangle and is 
parallel to a second side, then it bisects the third side. 

Proof: Assigned as an exercise* 



EXERCISES 103 

In Exercises 1-5, there is a triangle, &ABC t with points D and £ such that 
A-D-B and A-E-C, 75E | BC, and with lengths denoted as follows: 
p = AD, q = DB, rmAE,tm EC In each case, given three of the four 
numbers p. q, r, s, find the missing one. Draw and label a figure for each of 
these exercises, 

1. p = 5, q a 6, r = 75, * = [7] 

2. p = 8, q = G> r = 0, « = 9 

3. p = 18, <f = (TJ, r = 10, * = 13} 

4. p = [0, </ = »$, t = 7, * = 20 

5. p = V£ </ = "\A *" = 4 * = 

In Exercises 6-10, there is a triangle, A ABC, with points and E such tliat 
A-IKBaudA-E-C, and with lengths denoted as follows: x = AH,jj = AC T 
p — AD, q = DB, r = AE, tt = EC. In each case, given some of these 

lengths, determine whether or not the lines DE and BC are parallel. Draw 
a figure for each exercise. 



6. x = 10, 


if = is. 


p = 5, 


r=S 


7. * = 10, 


!/ = 15, 


p = 5, 


r=7.5 


8. p = 25, 


q = i5, 


r = 60, 


* = 3fl 


9. p a 0.9, 


s = 0,88, 


r = 0,81. 


q = 0.81 


10. p = ^ 


* = 2, 


r=2, 


t/ = 4 



11. Write a "reason" for each of steps 1 through 7 in the proof of Theorem 
10.8, 



10.5 Some Length ProportDomlitit* 413 

12. Coplan.tr and parallel lines a, b, c r d arc cut by transversals s and f sis 

•ii i'-."/-' ; Mn the figure, Given Lengths C& -> cements SS labeled in Lite lim- 
ine, find x and tj. 



_^ 



■fro 



-M 



5.4 



-¥d 



13. If tlirer distinct c:oplanar and parallel lines arc cut by two distinct par- 
allel transversals, then the lengths of the segments formed on one trans- 
versal are proportional to the lengths of the segments formed on the 
other transversal. What is the constant of proportionality in this case? 

14. In the proof of Theorem 10.8, Figure 10-7 suggests that s and / do not 
intersect in the portion of the plane between lines / and n. Draw a figure 
for this theorem which shows ,v and * intersecting at a point between 
lines / and n. b the proof for this theorem, as given, applicable for the 
case suggested by your figure? 

15. Consider again Figure 10-7. Suppose that the figure is modified to show 
s and f intersecting at a point P on the opposite side of hne / from it 
Using Theorem 10.6 and Figure 10-7 suitably modified and labeled* 
obtain some proportionalities involving lengths of segments with P as 
one endpoint, Use these proportionalities to prove that 

(AB, AQ = (DE, DP). 

16. Complete the following proof of Corollary' 10.8.1. Let A ABC be given 
with D the midpoint of AB and let I he the line through D, parallel to 

BC, and Intersecting A Cat E as shown in die figure. Let l\ be the line 

through A and parallel to J. Then l x | BC. Why? Then 

{AD t DB) = (AE, EC). Why? 

It follows that AD_ _ AE _ , .„. „ 

DB ~ EC~ ' 

Complete the proof by showing that / bisects AG. 

A 



4 — 


d/ 


\e 


Hi 











1 



414 



Similarity 






Chapter 10 


17. 


(An in 
of con 
length 
tractoi 
Copy 


formal geometry exercise.) With the aid of a rale 
i passes, draw two triangles, A ABC and AA'B'C 
of the sides (in centimeters) are as listed in the table. 
; measure the angles A, B, C, A', W, C to the n© 
the table and complete it by recording the angle in 


r and a pair 
so that the 
Using a pro- 
test degree. 

fiisurev 




AABC 


AA'B'C 




AB = 9.0 


A'W = 13.5 






BC = 7.0 


ffC = 10.5 








AC = 6,0 


A'C = 9.0 








m£A = [£J 
mlB = [?J 
roZC = [T| 


m/.A' = g] 
mLW m E3 
mlC = |1] 





1& {An informal geometry exercise.) With the aid of a ruler and protrac- 
tor, draw two triangles, A ABC and AA'B'C, with side lengths (in 
centimeters) and angle measures as indicated in the table. Measure the 
remaining parts of the two triangles and record the results. Measure 
lengths to the nearest 0,1 cm. aud angles to the nearest degree. 

AABC AA'B'C 



AB = 8.0 A'B' = 10.0 

BC = 10.0 JTC = 12.5 

AC = CO A'C = [7j 

mlA = {jl m£A' = {2] 

mZB = 46 m/.B' = i6 

i«ZC = (T) inZC = [TJ 



lfc {An informal geometry exercise.) With the aid of a ruler and a pro- 
tractor, draw two triangles* A ABC and AA'B'C with side lengths (in 
centimeters} and angle measures as indicated in the table. Measure the 
remaining parts of the two triangles and record the results. Measure 
lengths to the nearest 0.1 cm. and angles to the nearest degree. 



AABC AA'B'C' 



AB = 10 A'B' = 6 

BC = CD B'C = \T] 

AC=\?] AC = 

mZA = 32 mZA' = 32 

mZB = 51 mLW = 51 

mlC=\2} mZC = [?3 



10. 6 Triangle Similarity Theorems 41 5 



10.6 TRIANGLE SIMILARITY THEOREMS 

In our study of congruence for triangles we first defined congru- 
ence so that, by definition, all six parts of one triangle must be congru- 
ent to the corresponding parts of a second triangle in order for the 
triangles to lie congruent to each other. On the basis of our experi- 
ences with triangles it seemed reasonable to expect all six of the re- 
quired congruences involving sides and angles to be satisfied if certain 
sets of three of them are satisfied So we adopted the well-known Tri- 
angle Congruence Postulates, referred to as S.A.S., A,S,A., and S.S.S, 
Similarly, our experiences with triangles, especially the triangles of 
Exercises 17, 18, 19 of Exercises 10,5, suggest that, if certain combina- 
tions of some of the definitional requirements for a triangle similarity 
are verified, then all of the requirements for a similarity are satisfied. 
Since we adopted Congruence Postulates, it would seem reasonable to 
adopt Similarity Postulates. It turns out, however, that it is not difficult 
to prove what we want to know about similarity; hence in this instance 
postulates are not necessary. First, we prove a theorem that is useful in 
proving the main Similarity Theorems. 

THEOREM 10.9 If A ABC is any triangle and k is any positive 
number, then there is a triangle AA'B'C such that AA'B'C *— 
A ABC with constant of proportionality k. 

Proof: Let triangle A ABC and a positive number k be given. We 
consider three cases. 

Case 1. k <C \. 
Case 2. k = L 
Case 3. k> 1. 

We shall prove the theorem for Case 1 and assign the other two 
Cases in the Exercises. a 

Suppose that k < 1; then there Av 

is a point D on AB such that / ^s^ 

AD = k* AB and a point E on p/— — .--J>. 
£C such that AE = k*AC In / / ^\ 

Figure 10-8, DE is drawn so that g f -^5, 

k appears to be about 0,6 , Figure io-8 

It follows from the converse of the Triangle Proportionality 
Theorem that DE |, EC, It follows from theorems regarding parallel 
lines that 



LADE & L B and LA2ED ss L C. 



416 Staiilirfty Chapter 10 

Of course. LA^ I A. As you might expect, it is AADE that qualifies 
as it suitable AA'FC, that is, ADE «— * ABC is a similarity. So far 
we have shown that corresponding angles are congruent and, from the 
way we have chosen D and E t we know that 

(AD, AE) = (AB y AC). 

We need to show that 

{AD, AE, DE) = (AB, AC. BC). 

_Let F be the point of BC in which the line through £ and parallel 
toJB intersects BC. Then DE = BF(Why?) and 

{AE, BF) y (AC t BQ. Why? 
Substituting, we get 

(AE, DE) = (AC\ BQ. 
From tins proportion and the preceding proportion, 

(AD,AE)j(AB s Aq, 
it follows that 

(AD, DE, AE) = (AB t BC, AC), 
which completes the proof for Case 1 in which k < 1. 

THEOREM 10.10 (S,S.S. Similarity Theorem) Given A ABC 
and ADEF, if 

(AB, BC, CA) = (DE, EF, FD), 

then A ABC— ADEF. 

Proof; Let A A BC and A DEF such that 

(AB, BC, CA) = [DE, EF, FD) 

he given. (Sec Figure 10-9.) Suppose the constant of proportionality 
is k. From Theorem 10.9 it follows that there is a triangle AD'E'F such 
that AD'E'F *— ADEF Witt) proportionality constant ft. Then 

(AB. BC, CA) = (DE. EF, FD) with proportionality con- 

stant k. 

(&E\ FF, FD-) = (DE, EF, FD) with proportionality con- 
stant k, 

(AB,BC,CA) = (D , E\ET,F'D f ) with proportionality con- 
stant 1. Whv? 



10.6 Triangle Similarity Theorems 417 






D* *F D' 

Figure 10-9 

From this we conclude that AB = DT', BC = E'F, and CA = 
Ft/. It follows from the S.S.S. Congruence Postulate that 

AABC== AD'E'F, 

BecaU now that triangle congruence is u special case of triangle simi- 
larity and that triangle similarity is an equivalence relation. There- 
fore AABC- AD'E'F. But AD'IT-ADEF, it follows thai 
A ABC— AD£F, 

THEOREM 10.11 (S.A.S. Similarity Theorem) Given AABC 
and ADEF, if 

Z A =s LD and (AB, AQ = (DE, DF), 

then AABC- ADEF. 

Proof: Let AABC and ADEF be given with L A ss Z D and 

(AB, AQ = (D£, DF). 

(Use Figure 10-9 again.) Suppose die constant of proportionality is k. 
Let AD'E'F be a triangle such that AD'F/F ~ ADEF with propor- 
tionality constant k. Then 

AB = fc-DE, D'£' = Jfc«D£ t AB = WB % 
and 

AC? = k ■ DF, DT = Jt ■ DF t AC = D'F. 

It follows from the S.A.S. Congruence Postulate that 

AABC = AD'E'F. 

Then AABC ~ AD'E'F and AUWF ~ ADEF, and wc may con- 
clude that AABC - ADEF. 

COROLLARY 10.11.1 A segment which joins the midpoints of 
two sides of a triangle is parallel to the third side and its length is 
half the length of the third side. 

Proof: Assigned as an exercise. 



and 


AA=*AD f ; 


and 


IB^AF/- 


and 


AB = D'E'. 



418 Similarity Chapter 10 

THEOREM 10.12 (A.A. Similarity Theorem) Given A ABC and 
ADEF, if A A s Z D and ZB ss Z E, then AABC — AJ9£F. 

Proofi Let AABC and ADKFsuch liiat LA m AD and AB~z AE 
he given. (Use Figure 10-9 once more.) Let 

AB _ k 

Let AD'E'F be a Iriangle such thai AD'E'F -* ADEF with propor- 
tionality constant rf, Then 

AA^ID, ADsiAD\ 

ABmAM, AEatAE\ 

AS = k-DE, DT = k-DE, 

It follows from the A.S.A. Congruence Postulate that 

AABCs ADTF. 

Then A ABC - AD'E'F and A£>E'F - ADEF, and we conclude 
that A ABC — ADEF. 

Note that we have an S.S.S. Congnience Postulate and an S.S-S. 
Similarity Theorem f and that we have an S.A.S. Congruence Postulate 
and an S.A.S. Similarity Theorem, hut that we do not have an A.S.A. 
Similarity "Theorem to match our A.S.A. Congruence Postulate. Of 
course, we could, if wc wished, call our A.A. Similarity Theorem die 
A.S.A. Similarity Theorem. But if Z A a* AD and AB =* AE, we do 
not need to be concerned, about whether "AB is proportional to DE," 
Indeed, if AB and DE are any two positive numbers whatsoever, there 
is a number k such that 

AB-a It -JOB. 

Look at the tables you prepared for Exercises 17, 18, 19 of Exercises 
10.5. Do the measurement data recorded in the tables illustrate the 
triangle Similarity Theorems? They should. Which theorem does Excr- 
eta 17 illustrate? Which theorem docs Exercise 18 illustrate? Which 
theorem does Exercise UJ illustrate? 

We have written the triangle Similarity Theorems using quite a few 
symbols. Is it possible to state them in a more relaxed form without 
symbols? In the following versions of the theorems we use the word 
^corresponding" without "pinning it down." It should be understood 
in each case that a correspondence between the vertices of one triangle 
and the vertices of the other triangle is fixed so that there are corre- 
sponding parts. 



10.6 Triangle Similarity Theorems 419 

THEOREM 10.10 (S.S.S. Similarity Theorem— Alternate Form) 
If the lengths of the sides of one triangle are proportional to the 
lengths of the corresponding site of the other triangle, then the 
triangles are similar. 

THEOREM 10.11 (S.A.S. Similarity Theorem— Alternate Form) 

Tf an angle of one triangle is congruent to an angle of another tri- 
angle and if the lengths of die including sides are proportional to 
the lengths of the corrcspondint; sit Irs in the oilier triangle, then 
the triangles are similar. 

THEOREM 10J2 {A. A. Similarity Theorem— Alternate Form) 

If two angles of one triangle arc congruent to the corresponding 
angles of another triangle, then the triangles are similar. 

The following theorem points out that if the sides of one triangle 
are parallel to the sides of a second triangle, then the two triangles are 
similar. The property of parallel sides is a sufficient condition to ensure 
similarity* but, of course, it is not a necessary condition. 

THEOREM10.13 U triangles AABC and ADEF are such that 
A~B .1 m. BC I! EF, UA II FD, then AABC - ADEF. 

Proof: Assigned as an exercise. 

EXERCISES 10.6 

1. Prove Theorem 10,9 for the case in which k = L 

2. Prove Theorem 10.9 for the case in which k > 1. 



In Exercises 3 and 4. two triangles and the lengths of their sides are given 
by means of a labeled figure. 

a b AABC — APQR? 
Is AABC ~ AQPR? 
h AABC- APRQ? 
Is AABC— ARPQ? 
Is AACB ~ ARQF? 
Is ACBA - AQPR? 

4. Is AABC— A CDE? 
h AABC— ADEC? 
Is AABC- AEDC? 
Is A CAB- A DEC? 
Is AC/iA — ADCEr 
Is ABAC— A DEC? 




420 Similarity 



Chapter 10 



In Kxerciscs 5 and 6, two triangles arc given in a figure wilh some segment 

lengths and angle measures. 



5. 



6. 



Is AABC 
Is AABC 
Is AABC 
Is AABC 
Is Aj\BC 
Is AABC 



ADEF? 

A EFD? 
ADFE? 
AEDF? 
AFED? 
AFDE? 



Is AABC ~ ADEB? 
Is AABC— ADKE? 
Is A A KG— ABED? 
Is AABC— &JHBEP 
Is AABC— AEBD? 
Is AABC ~ AEDB? 




7. Given isosceles AABC with AB ~ AC* and with points D, /£, Fsueh 
dnit A-D-B, JJ-E-C, C-F-A, DE±AB, FE 1 XC, prove that 
ABD£— ^C3^. 

8, If at a certain Lime, in a certain place, a certain tree casts a shadow 40 ft, 

long and a 6-ft, man casts a shadow 2 ft. and 3 in. long, find the height 
of the tree. 



Exercises 9-11 refer to the figure with A-E-C, B-D-C, ED || A~B, and seg- 
ment lengths as marked. 

9. Name a pair of similar 
triangles and explain 
why they are similar. 

10, Find r. 

11, Find* 



12, In the figure, A-D-C and 
LABC^lBDC. Name a 
pair of similar triangles and 
explain why they are similar. 




10.6 Triangte Similarity Theorems 421 
D C 




13. In the figure* AB || CD, 
IDE X AC, W±7kC t A-E-F, 
and E-F-C. Name a pair of 
similar triangles and explain why 

they are similar. 



Exercises 14-20 refer to the figure with BTJ ,| OE, DB || EC, A-B-D, D-E-F, 
A-C-F f and A~F X F25. 

14. Prove AAFD ~~ &CFE. 

15. Prove ACF£ — AACB, 

16. Prove AAFD — AACB, 

17. If CF = 2, BD = 3, and 
AC = 8, find AB. 

18. If AB = 12, BD = 3, and 
BC = 8, find DF 

19* If AD = 18, AF = 9, and AC = 7, find EC. 
20. If CF = 3, KF = 4, and AC = 7, find DF. 




21. In the figure, ^fE and CD are altitudes 
of AABC, A-F-E, and C-F-D. 

(a) Prove AB£A - A.BDC. 

(b) Prove AAW- ACEF. 



22. Given AAJ3C with points D and 
£ such that A-D-B, B-E-Q 
AE X BC, CD X SB, prove that 
(AEtEfy-iCDrDB). 



23. Given two right triangles, A ABC and 
AAPQ, as in die figure, copy and com- 
plete the following proportionality involv- 
ing lengths of the sides of these triangles: 

{AB.BC,CA) = (AP T [2],[I]). 




24. Given AABC - ADEF, AB = §, BC = 7, AC = 10. DE = 7, fi 
EF and DF 



422 Similarity 



Chapter 10 




25. Given A ABC - APQH prove dial if A ABC is a right triangle, then 
APQR is also a right triangle, 

26. Prove Corollary 10.11.1. 

27. CHALLENGE PROBLEM, CiVCIl 

parallelogram ABCD with 
B-E-C, AE and ED intersect- 
ing at h\ and BF = $-BD, 
prove that BE = £ - BC 

28. Prove Theorem 10.13. Consider two cases: (a) AB and DE are parallel, 

BC and EF are parallel, CA and VD are parallel, and j b) AM and 15 JS are 

antiparailcl, BC and /Li*' are antiparaUel, CA and F£) are an ti parallel Use 
Theorems 7.26 and 7.28. 

10.7 SIMILARITIES IN RIGHT TRIANGLES 

Sometimes base and altitude are interpreted as segments and some- 
times as numbers (lengths of segments). In our next theorem altitudes 
are segments If AABC ~ AA'B'C, then we have agreed that A and 
A' are corresponding vertices, KB and A'B' are corresponding sides, 
and so on. It is natural to extend this idea to include corresponding 
altitudes, that is, altitudes from corresponding vertices. 

THEOREM 10.14 If two triangles are similar, then the lengths of 
any two corresponding altitudes are proportional to the lengths of 
any two corresponding sides, 

Proof: Given A ABC ~ AA'B'U, let D and D' be the feet of the 
altitudes from A to E€ and from A' to B'C, respectively. Let a = BC, 
b = CA, c = AB, h = AD, a' = B'C, b' = CA\ c' = A'B', 
h' = A'D'. 

We shall prove the theorem for the case in which B-D-C, as shown 
in Figure 10-10, The remainder of the proof is assigned in the Exercises . 



>1^ 




Figure 10-10 



10,7 Similarities In Right Triangles 423 

Since B-D-C, Z B and Z C are acute angles. Why? Then Z B' and 
Z C are acute angles. Why? Then it is impossible for D' to be either 
the point £' or the point C\ Why? Also, it is impossible that D'-B'-C 
or that W-C-D'. If ET-W-C, then AB'&A' is a right triangle with an 
acute exterior angle contrary to the Exterior Angle Theorem. There- 
fore B'-D-C as indicated in Figure 10.10, 

Now /_B ^ LB {Why?) and LBDA ^ LB'D'A* (Why?). It fol- 
lows from the A.A. Similarity Theorem that AADB — AA'D'B'. 
Therefore 



But 



(c. ft) = (c', A'). 



(a, 6, o) = (a', b f , c'l 



Therefore 

(a, c) == (a\ c') and (fo, cj = (//, 

It follows from the Alternation and Inversion Properties of Propor- 
tions that 

{K /i') f (c, v'% (c, c') = (a, a% and (c, c<) f (ft, I,*). 

It follows from the Equivalence Properties of Proportionalities that h 
and h' are proportional to the lengths of any two corresponding sides. 



THEOREM 10,15 If two triangles are similar, then their areas 
are proportional to the squares of the lengths of any two corre- 
sponding sides. 

Proof; Given A ABC ~ AA'iTC, let D and D' be the feet of the 
altitudes from A to W and from A' to B'C, respectively. Let a = BC, 
h = CA, c = AB r h = AD, a' = B'C. 1/ = CA\ tf = A'B\ h r = 
A'D'. (See Figure 10-11.) It follows from Theorem 10,14 that 



A 



B D 




B- ir 



Figure 10-11 



424 Similarity Chapter 10 

Suppose that h = ka; then h' = kif and 

\AABC\ = {ah = %a(ka) = $k)a* t 
| AA'B'C' 1 = \u'h' = \<f{W) = (JfttfA 

Therefore the areas of A ABC and AA'jB'C are proportional to a- unci 
(rt') 3 . Similarly, it may be shown that the areas are proportional to h 2 
and (by and'to c 2 and (c') a . 

THEOREM 10.16 In any right triangle the altitude to the hypote- 
nuse separates the right triangle into two triangles each similar to 
the original triangle, and hence also to each other. 

Proof: Let A ABC be a right triangle with the right angle at C and 

with D the foot of the altitude to the hypotenuse. Then D ,/- A and 
D /- B. (Which theorem is the basis for this assertion?) Also* it is iin 
possible to have D-A-B or A-B-D. (If either of these betweenness re- 
lations ig true, there is a triangle with D as one vertex with one interior 
angle a right angle and one exterior angle an acute angle. Which the- 
orem does this contradict?) Therefore D is an interior point of AB as 
suggested in Figure 10-12. 




Figure 10- IS 

You are asked in the Exercises at the end of this section to complete the 
proof by showing that A ABC, AACD, and I\CBD arc all similar to 
each other. 

Next we have two corollaries that follow easily from Theorem 
KXIG, but first we need some definitions. 



Definition 10.4 If P is a point and I is a line, the projection 
of P on Ms (1) the point P if F is on / and (2) the foot of the 

perpendicular from P to I if P is not on /. 

Definition 10.5 The projection of a set S on a line / is the 
set of all points Q on t such that each Q is the projection on J 
of some P in S. 



10.7 Similarities in Right Triangles 425 

Compare Definitions 10.4 and 10.5 witb Definitions 8,9 and 8.10 
in which we denned a projection on a plane. 

Note in Figure 10-12 that AD is the projection of AC on AB. In- 
deed, A is the projection of A T D is the projection of C, and every point 
Q such that A-Q-D is the projection of a point P such that A-P-C, 

Conversely, every point P such that A-P-C has as its projection on AB 
a point Q such that A-Q-D. 

Since the projection of AT on AB is a part of the hypotenuse in the 
situation of Theorem 10.16, we may say that AD is the projection of 
AC on the hypotenuse. Similarly, 7M is the projection of CB on the 
hypotenuse. 

COR OLLAR Y 10J0.1 The square of the length of the altitude to 
the hypotenuse of a rigfat triangle is equal to the product of the 
lengths of the projections of the legs on the hypotenuse. 

Proof; In Figure 10-12, AD is the projection of AC on AB, and UB 
is the projection of WC on AB. In the notation of the figure, we must 
prove that 

(CUP^AD-DB. 

Using Theorem 10.16 and some properties of proportionalities, wc 
have 

AACD - ACBD 

(AC, CD, AD) = (CB f BD, CD) 

{CD, AD) = (BD, CD) 

(CD)* = AD*DB 

COR OLL\R Y I 0. 1 &£ The square of the length of a I eg of a right 
triangle is equal to the product of the lengths of the hypotenuse and 
the projection of that leg on the hypotenuse. 

Proof- Assigned as an exercise. 

Definition 10.6 If a and b are positive numbers such that 

(a, x) = (*, b) 
or that 

fe *) f ft *). 

then % is called a geometric mean of a and h. 



426 Similarity 

Note that if 

or if 



Chapter 10 



(*. x) f fe b) 

(x, a) = (h, 4 



then x 2 = ab (Why?) and ar = y/aB or x = — yS5. We often call y/aB 
the geometric mean of a and b. In view of Definition 10,6, Corollary 
10.16 .1 and Corollary 10.16.2 can be restated as follows, 

COROLLARY 10.16.1 {Alternate Form) The length of the alti- 
tude to the hypotenuse of a right triangle is the geometric mean of 
the lengths of the projections of the legs on the hypotenuse. 

COROLLARY 10A6.2 (Alternate Form) The length of a leg of 
a right triangle is the geometric mean of the lengths of the hypote- 
nuse and the projection of that leg on the hypotenuse. 

In Chapter 9 we proved the Pythagorean Theorem using proper- 
ties of areas and suggested two other area proofs in the Exercises. One 
of the shortest proofs of the Pythagorean Theorem is an algebraic proof 
that follows easily from Corollary 10.16.2. We state the Pythagorean 
Theorem again and outline a proof that employs Corollary 10, 16.2, We 
also proved the Converse of the Pythagorean Theorem in Chapter 9. 
We state the converse again; we shall not prove it again. 

THEOREM 10,17 (TJte Pythagorean Theorem) In any right tri- 
angle the square of the length of the hypotenuse is equal to the sum 
of the squares of the lengths of the two legs. 

Proof: Let A ABC With a right angle at C be given. (See Figure 
10-13.) Let D he the foot of the altitude to the hypotenuse AT?. Ixt 
AR = c,BC = a, CA = b. AD a x t twdDB == c - *. Then it follows 
from Corollary 10.162, with a the length of a leg and c — x the length 
of its projection on the hypotenuse that a 2 = (c — x)c t and with b the 
length of a leg and x the length of its projection on the hypotenuse that 
b 2 =. xc .The proof may be completed by .showing thala? ■+ fr 2 = c 2 . 



Figure 10-13 



A 




* D 



10.7 Similarity* in Right Triangles 427 

THEOREM 10.18 {Converse of the Pythagorean Theorem) If 
a 2 + fc 2 = c 2 , where a, h, c are the lengths of the sides of a triangle, 
then the triangle is a right triangle with c the length of I he 

hypotenuse. 



EXKKCJSES 10.7 

I, Given right triangle A ABC with hypotenuse AB, let D he the foot of 
the altitude to EC. E the foot of the altitude to AC, and P the foot of the 
altitude to AB, How many distinct points are there in the set 

(A,B, C,«, E.F)? 

In Exercises 2-7, there is a figure show mga right triangle with hypotenuse 
AT3 and with D the foot of the altitude to aB, In each case, given the lengths 
of some of the six segments, find the lengths of the other segments. Express 
your answers in exact form using radicals if necessary. 




3. B 




C 4 



4, b 





428 Similarity Chapter 10 

H, Find the perimeter of an equilateral triangle if the length of each of its 

altitudes is 10. 
9. Find the length of the diagonal of a rectangular floor to the nearest foot 

if the floor is 21 ft. wide and 28 ft. long. 

10. A ladder 12 ft long reaches to a window sill on the side of a house. If 
the window sill is 9 ft. at»ove the (level} ground, how far is the foot of 
the ladder from the side of the house? 

11. Find the length of the hypotenuse of a right triangle if its legs each have 
length 1. Express the answer exactly using a radical. 

12. Find the length of the hypotenuse of a right triangle if its legs each have 
length 100. 

13. Find the lengths of the legs of an isosceles right triangle whose hypote- 
nuse has length 1. 

14. Find the lengths of the legs of an isosceles right triangle whose hypote- 
nuse has length 2. 

15. Find the lcugtlis of the legs of an isosceles right triangle if the length of 
its hypotenuse is \/2. 

16. If one leg and the hypotenuse of a right triangle have lengths 1 and 2, 
respectively, find the length of the other leg. 

17. Find the I ength of a leg of a right triangle if the other leg and the hypote- 
nuse have lengths 1 and \/3~, respectively. 

18. Find the length of the leg of a rigfrl triangle if the other leg and the hy- 
potenuse have lengths 100 and 100\/3~ t respectively. 

19. Given A ABC with m£ C - 90 and with D the midpoint of AS aiid£ 
the midpoint of EC, prove that ACED m ABED. 

20. For AA BC t m Z C = 90 and D is the midpoint oiA~B. If AC = y7 and 
BC = 3, find CD. 

21. Complete the proof of Theorem 10.16 by showing that 

ABC «— * ACD 
and 

ABC * — ► CBD 

are similarities. (See Figure 10-12.) It will then follow from the equiva- 
lence properties of similarity for triangles that ACD * — * CBD is also 
a similarity. 

22. Prove Corollary 10.16.2 fOT the leg AC in Figure 10-12. 

23. Prove Corollary 1Q.1&2 for the leg EC in Figure 10-12. 

24. See the proof of Theorem 10.17. Show that a 3 + fe* = c 2 . (See Figure 
10-13.} 

25. If A ABC ~ A DEF and 5 • A = 3 * DE, what is the ratio of the length 
of an altitude of the smaller triangle to the length of the corresponding 
altitude of the larger triangle? Which theorem justifies your answer? 

26. In Kxcrcise 25, what is the ratio of the area of the smaller triangle to the 
area of the larger triangle? Which theorem justifies your answer? 



10.8 Some Right Triangle Theorems 429 

27« If 16- \APQR\ = 25 ■ A ABC and If APQR — AAHG, what is the 
ratio of PR to AC? Which theorem justifies your answer? 

28. Prove Theorem 10.14 for the case in which D = B or D = C 

29, Prove Theorem 10.14 for the case in which D-B-C (the proof for the 

case in which B-C-D Is similar to the proof for the case in which 
D-B-C). 



10.8 SOME RIGHT TRIANGLE THEOREMS 

Following arc some theorems regarding right triangles. Although 
the) f are not profound, they arc useful theorems that every mathemat- 
ics student who has studied formal geometry ought to know. These 
theorems should not surprise you. If you worked the exercises in Ex- 
ercises 10.7, you will recognize them as "old stuff." 

THEOREM 10,19 The median to the hypotenuse of a right tri- 
angle is one-half as long as the hypotenuse. 

Proof; Let AABC be a right triangle 
with D the midpoint of the hypotenuse* 
{See Figure 10-14.) We want to prove 
that CD = \*AB, or equivalent! v, that 
CD = DB. I^et point E be the midpoint 
of CB. Then 




C B 

Figure 10*14 



(BD, BE) = (BA, BC) (Why?) and LB^LB. 

It follows from the S.A.S, Similarity Theorem that A ABC— A DUE. 
Then 

mLBED = mLBCA = 90 = mLCED. 

It follows from the S.A.S. Congruence Postulate that ACDE=* 
ABDE and therefore CD = DB. 

There are some special right triangles that are referred to in special 
ways. First we mention the 3, 4, 5 triangles. A triangle whose sides 
have lengths 3, 4 t 5 is a right triangle. We know this since 3* + 4* m 5 2 < 
(Arc we using the Pythagoreun Theorem when we muke tills em illu- 
sion, or are we using its converse?) 

Given a distance function, there are infinitely many 3, 4, 5 right 
triangles. Indeed, if A is any point in space (infinitely many choices 
here) and if B is any point such that AB = 5 (infinitely many choices 
here), there arc infinitely many possible points C so that A ABC is a 
right triangle with AB = 5, BC = 3 t and CA = 4, and infinitely many 



430 Similarity Chapter 10 

possible points C such that AB = 5, BC = 4, and CA — 3, But there 
are many, many more, not included among these, that are also fre- 
quently referred to as 3, 4, 5 right triangles as our next theorem 

suggests. 

THEOREM 10.2(1 (The 3, 4> S Theorem) If a: is any positive num- 
ber, then every triangle with side lengths 3x, 4x, 5.r is a right 
triangle. 

Proof: Let &ABC be a triangle with BC = 3, CA = 4, AB = 5. Let 
x be any positive number. Let &A'B'C f be any triangle with B'C = 3x, 
CA f =: 4s, A'B' = 5x. Then AA'B'C - AABC. (Which triangle 
Similarity Theorem do we use in making this deduction?) Since A ABC 
is a right triangle, it foDows that A A 'B'C is a right triangle, and this 
completes the proof. 

A triangle is called a 3, 4, 5 triangle if its sides arc of lengths 3, 4, 5 
or if the lengths of its sides are proportional to 3, 4, 5, AH 3,4, 5 triangles 
arc right triangles. 

THEOREM 10.21 (The 5, 12, i.1 Theorem) If x is a positive num. 
ber and if the lengths of the sides of a triangle are ox, 12x, and 13x, 
then the triangle is a right triangle. 

Proof: Assigned as an exercise. 

A triangle is called a 5, 12, 13 triangle if the lengths of its sides are 
proportional to 5, 12. 13. 

THEOREM 10.22 (The /, J, \/2 Theorem) If the lengdis of the 

sides of a triangle are proportional to 1. 1, y/S. then the triangle is 
an isosceles right triangle. 

Proof: Let the lengths of the sides of a triangle be o, fc, c and suppose 

{«, h c) = (1, 1, Vl). 

Then there is a positive number k such that a = k • 1, h = k • 1, and 
Gssk*y/£. Then 

a = b, 

t& + 6 s = fe 2 + Jt= = 2fc 2 , 

c 2 = (k^f = 2&, 
a 2 + fc 2 = cK 
Therefore the triangle is an isosceles right triangle. 



10,8 Some Right Triangle Theorems 431 

A triangle is called a I, 1, V5 triangle, or a 45, 45, 90 triangle, if 

the lengths of its sides are proportional to 1 , 1 , -\/S, 

THEOREM 10.23 (The 1, y/3, 2 Iheorem) If the lengths of the 
sides of a triangle are proportiona] to 1 1 \/3, 2, then it is a right tri- 
angle with its shortest side half as long as its hypotenuse. 

Proof: Assigned as an exercise. 

A triangle is culled a 1, \/S, 2 triangle if the lengths of its sides are 
proportional to 1, y/3 t 2, 

A triangle is called a 30, 60, 90 triangle if the measures of it* acute 
angles are 30 and 60. 

THEOREM 10,24 A triangle is a 30, 60, 90 triangle if and only 
if it is a 1, \/3, 2 triangle with the shortest side opposite the 30 de- 
gree angle. 

Proof: Let A A BC be a 1 , \/3 f 2 triangle an d k a positive mm Vie r such 
that AC = fc, BC = s/Sk, and AB = 2k. (See Figure 10-15.) 



Let D he the point on opp CA such that CD — k. Then 

1. AABC=z ADBC 1. Why? 

2. AB = DB = DA 2, Why? 

3. mlABD + mlBDA + mlDAB = 180 3. Why? 

4. mlABD = mlBDA = m/.DAB = 60 4. Why? 

5. ml ABC = mlDBC 5. Why? 

6. ml ABC + mlDBC = 60 6. Why? 

7. ml. ABC =30 7. Why? 

8. ml8AC = 60 8. Why? 

9. mlACB = 90 9. Wiry? 

Since L ABC is opposite the shortest side of A ABC, this completes the 
"if* part of the proof. 



432 Similarity 



Chapter 10 



Suppose next that A ABC is a 30, 60, 90 triangle. (See Figure 
10-16.) Let D be the unique point on opp CB such that CB = CD, 




Figure 10-lfl 



Draw DA. Then 



10. AABC s AADC 

11. m/LCAB = mZ. CAD = 30 

12. mLBAD as mZADB = mlDBA = GO 

13. BA = AD = DB 

14. BC = CD 

15. BC + CD = BD 

16. 2BC a AB 

Let BC = fc. Then 

17. AB = 2k 

18. (AQ* + (BC) 2 = (AB)* 

19. (Aqz + A' 2 = 4*2 

20. (AC)* = 3Jfc* 

21. AC= VSJfc 

22. (BC,CA.AB) = (1, x/3',2) 



10. Why? 

11. Why? 

12. Why? 

13. Why? 

14. Why? 

15. Why? 

16. Why? 



17. Why? 

18. Why? 

19. Why? 

20. Why? 

21. Why? 

22. Whv? 



This shows that A ABC is a 1, \/5, 2 triangle and hence the "only if* 
part of the proof is completed. 



Note that in some of these names for special triangles the numbers 
are side lengths (or numbers proportional to them), whereas in others 
they are angle measures. There should be no confusion in regard to the 
30, 60, 90 name and the 45 f 45, 90 name. Because 30, 60, 90 are not 
the lengths of the sides of any triangle, and 45, 45, 90 are not the 
lengths of the sides of any triangle. Which postulate justifies this 
statement? 



L0.8 Some Right Triangle Theorems 433 



EXERCISES 10.8 

L If AABC is a right triangle with ml C = 90, AC = 60, BC - 80, and 
with D the midpoint of AB. find CD. 

2, (See Figure 10-14) Z ACS and Z DEB are congruent angles in the sit- 
uation represented by this figure. Find several other pairs of congruent 
angles. (Six more pairs would be rather good.) 

3. In the proof of Theorem 10,19 we asserted that ACD£a ABDE. 
Write a two-column proof for this deduction. 

4* In a book on the history of mathematics find something about the rope 
stretchers in ancient Egypt Explain the connection between rope 

stretchers and right triangles, 

5. A baseball diamond is u square whose sides are 90 ft. long. What is the 
distance (to the nearest foot) between first and third base*? 

6. The figure represents a cube whose six faces are 1 in. by 1 in. squares. 
Using the Pythagorean Theorem twice, once on ABCD and once on 
AABD, find AD, Express the answer exactly using a radical if neces- 
sary. (Why is LABD a right angle?) 




7. A room in the shape of a rectangular box is 15 ft wide, 1 8 ft. long, and 
8 ft. high. Find the distance to the nearest foot between one corner of 
the floor and a diagonally opposite comer of the ceiling. 

In Exercises 8-16, the lengths of the hypotenuse and one leg t>i' a right tri- 
angle are given. In each exercise, the triangle is a 1. l t \/2 triangle, or a 
1, \/3» 2 triangle, or a 3, 4, 5 triangle, or a 5, 12, 13 triangle. Determine 
which one. 



8. 100,50 

9. 100,60 
10. 100. 50 VI 



11, 100,80 

12, ]00, 50\/3 

13, NX), 92^ 



14. \/^,iv^ 

15. 145, 116 

16. 63, 25 



434 Similarity 



Chapter 10 



17. 



I .et^ A ABC be a right triangle with the right angle at C The midpoint 
of AB is the center of a circle which lies in the plane of the triangle and 
which contains the points A and B. Does the point C lie inside of the 
circle, on the circle, or outside of the circle? Why? 
The figure suggests a point A on a high bluff above a level plane. If the 
angle of elevation of the point A from the point C is n 90 degree angle 
and if it is 200 ft. from C to B t wliat is the height BA to the nearest 10 ft,? 
Assume that / CBA is a right angle. 




-::iC^S, :*>"" ~ 



19, The figure represents an observer A in an airplane 5000 ft. directly 
above a point 8 on the ground. If J3 T C T D arc three colllnear points on 
the ground and if m L ABC = 90, m L BAC = 45, m L CAD s 15, find 
to the nearest 100 ft. the distance from C to D. 




20. Find three positive integers, a t b t c such that \/a, \/' , » V^&re the 
lengths of the sides of a right triangle. How many such triples of positive 
integers arc there? 

21. If rand y are any positive integers, distinct or not, show that y/x, \/y, 
\fx + tj are the lengths of the sides of a right triangle. 

22. In the figure is shown a right triangle A ABC with CD the altitude to 
the hypotenuse. If AD = 1, DC = r, BD = y, show that * = y/y. 




10,8 Some Right Triangle Theorems 435 

23. In the figure below, ABCD is a parallelogram with AB = 76, AD m 50, 
m£A = 30, and h the length of the altitude from D to AS, Find 

\ABCD\. 




24. A parallelogram has adjacent sides of lengths 22 and 14. If the measure 
of one of its angles is 30. find the area of the parallelogram. 

25. Find the area of a rhombus of side length 12 if one of its angles has a 
measure of 60, 

2& The measure of each base angle of an isosceles triangle is 30 and each 
of the two congruent sides has length 24. 

(a) How long is the base? 

(b) What is the area of die triangle? 

27. In the figure. ABCD is a trapezoid with AB \TJD, AD = BC = 20, 
CD = 28, and ml A = mlH = GO. Find \ABCD\. 





Use the figure to complete the proof that in a 30, fiO right triangle the 
side Opposite the 30 degree angle is one-half as long as the hypotenuse. 




Proof: In the figure, ml A = 30, m L B = 60, and D is the midpoint 
of AB. Show that A BCD i,s equilateral and that 



BC = CD a 1 • AB. 



2d. In a 30. 60 right triangle, the length of the hypotenuse is 8 V^- 

(a) Find the length of the shorter leg. 

(b) Find the length of the longer leg. 

(c) Find the area of the triangle. 
30, Prove Theorem 10.21. 



436 Similarity Chapter 10 

31. Prove Theorem 10.23. 

32. If u and c are positive integers such that u > v, and if A ? B t C arc points 
such dial AC = 2av, BC = u* - p 2 , AB = u* + &, prove that &ARC 
is a right triangle. (This exercise also appears in Chapter 9, but it is good 
for a repeat appearance here.) 

33. See Exercise 32, If u uiid t: are relatively prime positive integers (this 
means tluit no positive integer except 1 divides both of them), if it 
and o are not both odd, and if u > c, then it can be shown that the three 
integers, u 2 — c 2 , 2uc„ i* 2 + v' £ , are relatively prime. If the lengths of the 
sides of a light triangle are relatively prime positive integers, the tri- 
angle is called a primitive Pythagorean triangle and the triple of its 
side lengths is called a primitive Pythagorean triple. Two examples of 
primitive Pythagorean triples are '(3,4,5) and (5,12,13). Find five 
more primitive Pythagorean triples, 

34. challenge problem, TTie figure shows a right triangle A ABC with 
CD the altitude to the hypotenuse. 




If AC = p, BC = q t CD = f, prove that Jj- + i = \. 



CHAPTER SUMMARY 

The central theme of this chapter is SIMILARITY. The concept of 
similarity is based on our experiences with objects which have the same 
shape. The relationship of lengths in one figure to the corresponding lengths 
in a similar figure suggests the idea of a PROPORTIONALITY. In this chap- 
ter we studied the properties of proportionalities and we used them in de- 
veloping the geometry of similar pjlygons. 

The key theorems of this chapter include THE TRIANGLE PROPOR- 
TIONALITY THEOREM. THE CONVERSE OF THE TRIANGLE PRO 
PORTIONALITY THEOREM, THE &&& SIMILARITY THEOREM, 
THE S.A.S, SIMILARITY THEOREM, THE A.A. SIMILARITY THE- 
OREM, THE PYTHAGOREAN THEOREM, and THE CONVERSE OF 
THE PYTHAGOREAN THEOREM. 

The chapter concludes with a study of special right triangles. A knowl- 
edge of these triangles will prove useful as you continue your study of 
mathematics* 



Review Exercises 437 

REVIEW EXERCISES 

In Exercises 1-1 0, complete the statement so that it will be a proportionality. 

1.(5.12) = (35,0) 

2. (1,2,3,4, 5) = (0,0 EJ. Q 15) 

3. (25, 60, 65) = (Q], 12,0) 

4. a io f 2i) = (a. mm) 

5. (4. 10, 21) = ([3,6, HI) 

6. (4, 10,21)= (0, 0,6) 

7. (5000, ,3000, 1500) = (200, 0, 0) 

MiOO,400,5W) = (II],m,10) 
9. (27, 27,81) = (a [7], 3) 

10. (357, 1309, 833) = ([?}, 11, [7J) 

In Exercises 1 1-20, determine If the given statement is true or if it is false. 

11. If x = y, then (5, x) - (5, y). 

12. lfxyt=y, then (5, x) = (5, ij). 

14If f = f'**>(3.3)f (*>!/>> 

15. If^^^.aieniAx)^^/ 

16. If x* = a&. then (a> x) = (*, £>), 

17. If x* = ab, then (*, a) = (b, x). 

18. If («. fc) = (c, <f), then (a. h) = [d, c), 

19. If (a, fc) = (c, a*), then (a, <:} = (fe, rf). 

20. If acf = hc t then (n, &) = (c, 5). 

21. If AABC is any triangle, then AABC — AABC. Which property of 
an equivalence relation docs this illustrate: the Reflexive Property, the 
Symmetric Property, or the Transitive Property? 

22. If A ABC- A DICE then &DEF ~~ A ABC. Which property of an 
equivalence relation does this illustrate? 

23. If AABC- ADKF and ADEF — AG//I, then A ABC— ACHL 
Which property of an equivalence relation does this illustrate? 

24. State the Triangle Proportionality Theorem and show how it may be 
used to prove the theorem regarding the lengths of segments formed by 
two transversals cutting three or more coplanar and parallel lines. 

25. State the three triangle Similarity Theorems. 



438 Similarity Chapter 10 

2& According to Theorem 10,9, if A ABC is any triangle and k is any posi- 
tive number, then there is a triangle &DEF such that ADEF= A ABC 
with proportionality constant t. F.xplain how tills theorem was used in 
proving the triangle Similarity Theorems, 

27. Explain why there is an A. A, Similarity Theorem but no A.5.A. Simi- 
larity Theorem. 

2S. Prove that the altitude to the hypotenuse of a right triangle determines 
two triangles that are similar to each other. 

29. Using similar triangles, prove the Pythagorean Theorem. 

30. If A, B> C, D are points such that A-D-B, AD = BD = 25, AC = 30, 
BC = 40, find CD, 

31. If the lengths of two sides of a right triangle are 10 and 15, find the 
length of the third side. (Two possibilities.) 

32. Find the measures of the angles of a triangle if the lengths of its sides 
are \/2~. V? 2 - 

33. Find the measures of the angles of a triangle if the lengths of its sides 
are \/S, 2\/3~.3- 

34. We know that SO, 60, 90 cannot be the lengths of the sides of a triangle. 
Which postulate justifies this statement? 

■ In Exercises 35-43, refer to Figure 10-17 in which A ABC is a right triangle 
with the right angle at C, CD is the altitude from C to the hypotenuse, 
AD = x, DB = «, AB = c,BC - a, AC = h, and CD = h. 




Figure 10-17 

35. A ACS — A ADC and A ACS ~- A[TJ. By the symmetric and prop- 
erties of similarity for triangles, A[T] — A(7J. 

36. h 2 equals the product of x and |T] (o, b t or tj). 

37. h 2 equals the product of x and |T| {a, c, or tj), 
38* a 2 equals the product of y and \T\ (k c > or x}> 

39. If x = 16 and tj = 9, find h,a, and h. 

40. If D is the midpoint of AS, then CD = [T] (in terms of c), 

41. If m L A = 30 and c = 15, then a = [TJ. 

42. If a s 10 and c = 20, then b = [U and m£A = [?]. 

43. If ml A ^ 45 and a = 12, then b = JT] and c = [?]. 



Review Exercis«f 439 

44. Copy and complete: In a 30, 60 right triangle, the side opposite the 30 
degree angle is [?] (in terms of the hypotenuse). 

45. If a boy 5 ft, tall casts a shadow 2 ft. long, how high (to the nearest 10 ft.) 

is a tree if its shadow is 73 ft. long? What assumptions did you make in 
working this exercise? 

48. Find the distance from C to A& if AC = 10, BC = 10 V5, AB = 20, 

47. If the hypotenuse and a leg of a right triangle have lengths 241 and 230, 
respectively, find the length of the other leg. 

48. challenge froblem. Given rectangle ADEH and points B and 
C on AD such that HA = AB = BC = CD = DE = 1, prove that 
m/LEAD = mlEBD = mAECD. 





Chapter 




Joyce fi. Wtlsoii/Photo Researcher* 



Coordinates 
in a Plane 



11.1 INTRODUCTION 

In Chapter 3 we introduced the idea of a coordinate system on a 
line. Recall that if F and Q are any two distinct points* then there is a 

unique coordinate system on FQ with P as origin and Q as unit point. 
Tims a coordinate system on u line is determined by choosing any two 
distinct points on the line, one of them the origin and the other the \ I j ii I 
point If, on line Z, F is the origin and Q is the unit point, then FQ is 
called the unit segment for the coordinate system on / determined by 
P and Q. A coordinate system on I is a one-to-one correspondence be- 
tween the set of all points of I and the set of all real numbers. The 
numbers associated with the points of /are called coordinates, and they 
can be used to determine distances (in the system based on FQ as the 
unit segment) between points on I. 

In diis chapter we introduce the idea of a coordinate system in a 
plane. In a plane, each point is associated with a pair of numbers, rather 
than a single number. After proving some basic theorems concerning 
a coordinate system in a plane, we develop some equations for a line. 
We then show bow coordinates can be used to provide simpler proofs 
of some geometric theorems. 



4.42 Coordinates in a Plane 



Ctiapter 11 



H.2 A COORDINATE SYSTEM IN A PLANE 

Suppose that a plane is given and, unless we specify otherwise, that 
all sets of points under consideration are subsets of this plane. Suppose 
further than a unit segment is given and that all distances are relative 
to this unit segment unless otherwise indicated. 

Let OX and OY be perpendicular lines in the plane intersecting 

in the point O as shown in Figure 11-1. Let / and / be points on OX 

and Vi t respectively, such that 

Ol = OJ = 1. 

There is a unique coordinate system on OX with origin O arid unit 
point I, This is called the \ -coordinate system, and the coordinate of a 

point R of OX in this system is called the x-eoordiiuiteor abscissa of R. 

t Y 
6 
i 



-6 -s -4 -3-2-1 



-•1 



/, 



It 



X 



12 3 4 5 



Figme 11-1 

In Figure 11-1, the abscissa of R is 5. There is a unujue coordinate 

system on OY with origin and unit point J, This is (railed the 

y-eoordinate system, and the coordinate of a point S of OY in this sys- 
tem is called the y-coordinate or ordinateof S. In Figure 11-1, the ordi- 
nate of S is —2. Name the coordinates of Tand V in Figure 11-L Is it 
necessary to specify the coordinate system in each case? Why? 

The line OX is called the ■ mlnmrl OY is called the y-axiv Together 
they are called the coordinate axes. Their point of intersection, O. is 
called the origin and the plane is called the .vv-plimc. Although we 
usually represent a line with a segment and an arrowhead at each end, 
it is common practice to represent an axis with a segment having an 
arrowhead only on the end that "points in the positive direction." In 
many of the figures in this book axes are represented in this way. 



11.2 A Coordinate System in a Plan* 443 

The projection of a point P on a line f is (1) P itself if F is on /, and 
(2) the foot of the perpendicular from F to /if ? is not on /. (See Defini- 
tion 10.4.) Since the perpendicular segment from an externa! point to u 
Line is unique, each point in a plane has a unique projection on a given 
line in that plane. 

In Figure 11-2, PA is perpendicular to the x-axis at A and FB is 
perpendicular to the {/-axis at B. Therefore A is the projection of Fori 
the x-axis and B is the projection of P on the (/-axis. If 4 and 3 are the 
coordinates of A and B> respectively, then we call the ordered pah* of 
numbers (4, 3) the coordinates of F. 















. 


^ 


























s 

4 






















































* 






* 


















a 

i 


^ 




























-±A s 




P5-4-3-2-: 


I 2 i 'A 5 












- 






















: 






















































































. 


L.J 



















Figure 11-S 

More generally, if F is any point in the xy-planc, the x-coordinatc 
(abscissa) of F is the x-coordinate of the projection of F on the x-axis. 
The ^-coordinate (ordinate) of F is the (/-coordinate of the projection 
of F on the {/-axis. We call the x-coordinate of F and the incoordinate 
of F the coordinates of P. The xy-coordinates or, simply, the coordi- 
nates of F are an ordered pair of real numbers in which the abscissa is 
the first number of the pair and the ordinate is the second. Thus if the 
abscissa of F is a and the ordinate of F is 6, then the x {/-coordinates of 
P are written as (a, b). 

THEOREM 11,1 The correspondence which matches each point 
in the .riy-plane with its xy-coordinates is a one-to-one correspond- 
ence between the set of all ordered pairs of real numbers and the 
set of all points in the xy-plane. 

Proof: In the xt/-plane, let an x-coordinate system on the x-axis and a 
{/-coordinate system on the {/-axis be given. Let F be any point in the 

given ary-plane, If A and /? are the projections of f on the x-axis and the 



44* Coordinates in a Plane Chapter 11 

(f-axis, respectively, let the abscissa of A be a and the ordinate of B f*? 
k Since each point on the *-axis has a unique x-coordinate and each 
point on the y-axis has a unique (/-coordinate, and since the projections 
of P on the x- and (/-axes are unique, it follows thai there is exactly one 
ordered pair of real numbers (a, h) that corresponds to the point P. 

Conversely, let (a, b) be any ordered pair of real numbers; then 
tiiorc is a unique point A on the x-axis with abscissa a and a unique 
point B on the (/-axis with ordinate h Also, there is a unique line fa 
through A and perpendicular to the x-axis and a unique line fe through 
B and perpendicular to the (/-axis. These two Hues intersect (Why?) 
in a unique point P. Hence every ordered pair of real numbers corre- 
sponds to exactly one point in the given xt/-plane and the proof is 
complete. 

Definition 11.1 The one-to-one correspondence between 
the set of all points in an xi/-plane and the set of all ordered 
pairs of real numbers in which each point P in the plane cor- 
responds to the ordered pair [a, b), in which a is the x-coordi- 
nate of P and b is the y-coordinate of F, is an v l i: : mate 
system. 



Since there are many pairs of perpendicular lines in a plane and 
since any pair of such lines may serve as axes, it follows that there are 
many ^-coordinate systems in a given plane. In a given problem sit- 
uation, we arc free to choose whichever coordinate system seems most 
appropriate. 

In view of the one-to-one correspondence between the set of alt 
ordered pairs of real numbers and the set of all points in a given 
xf/-plane, it is clear that symbols used to denote the ordered pairs may 
be used to denote the corresponding points. Thus, if P is the point 
whose coordinates are the ordered pair (4. — 3) t we may speak of the 
point (4, —3) or we may write F— (4, —3). Sometimes we simply 
write P{4, -3). 

It should be noted that the numbers in an ordered pair need not 
be distinct. Thxis (3, 3) is an ordered pair of real numbers, Of course, 
the ordered pair (3, 5) is not the same as the ordered pair (5, 3). Indeed, 
(0, b) = {e. d ) if and only if a = c and h = d. 

As shown in Figure 1 1-3, it is customary to think of the unit point 1 

as lying to the "right" of the origin (that is, on ray OX in the figure) and 

of unit point 1 as lying "above" the origin (that is, on ray OY). This 
means, then, tin at the points on the x-axis with positive abscissas lie to 
the right of the origin and those points on the x-axfs with negative 



1 1*2 A Coordinate System In a Plant 445 











A 


'V 




































































































rS 




















*? 








x i 




-3 















5 











































ngmm 1 t<3 



abscissas lie to the "left" of the origin (that is., on opp OX). Where do 
the points on the y-axis with positive ordinate* lie? Where do the points 
on the y-axis with negative ordiriates lie? 

There are situations in which it is convenient to think of the posi- 
tive part of the i-axis as extending to the left, or the positive part of 
the y-axis as extending downward, or some other variation. However, 
it will not be necessary to do this. 

A line in a plane separates the points of the plane not on the line 
into two half planes, the line being the edge of each halfplane. Simi- 
larly, the coordinate axes separate the points of an xy-plune not on the 
axes into four "quarter-planes," or quadrants, the union of whose edges 
is the axes. For convenience, these quadrants are numbered 1, 1 J, ITT, IV 
as indicated in Figure 1 1 -4 . 



+y 



a 



*<0 

y>0 




x>Q 
y>0 


x \ 











in 




IV 




x<0 




*>0 

>'<0 





Figure li-i 

Quadrant I is the set of all points {x t tj) such that % > and y > 0. 
Quadrant II is the set of all points {x t y) such that x < and y > 0, 
Quadrant III is the set of all points {x, y) such that X < and y < 0, 
Quadrant IV is the set of all points (r. y) such that x > and y < 0. 



446 Coordinates in a Plane Chapter 11 

We can describe the coordinates of those points (x, y) on OX by 

x > and y = 0, and the coordinates of those points on opp OX by 
x < and y = 0. Describe, in a similar way, the coordinates of those 

points (i% y) on OY: on opp OY, 

Since we usually think of an xi/-eoordinate system oriented as we 
have shown in Figures 11-1 through 11^1, it is customary to call all lines 

parallel to OY vertical lines. Similarly, we cal all lines parallel to OX 
horizontal lines. 

It is often convenient to use **above," "below," "right," "left" to 
describe the position of a point. However, we can get along without 
these words if challenged to do so. For example, we could describe the 
position of the point P = (2, —5) by saying that P is 5 units "below" 
tlie x-axis and 2 units to the "right" of the y-axis in an xiy-pl&ne. Or we 
could say that V is in the fourth quadrant, that it is on a vortical line 
which intersects the x-axis in a point 2 units from the origin, and diat 
it is on a horizontal line which intersects the y-axis in a point 5 units 
from the origin. 

EXERCISES 11.2 

■ In Exercises 1-S, name the quadrant in which the paint lies. 

h (-2,4) 5. {*, -2) 

1 (7, -3) 6. (-7.3, -1) 

& (-V2, -5) 7. (-V5, V3) 

•I) 



4. (1.2, 6) 8. & -I) 



In Exercises 9-18, describe the set of all points (x, y) which satisfy the given 
conditions. 

9. x <C 0, y < 14. x is any real number, y = 4 

10. x > 0, y < 15. x = —2, y is any real number 

1L x < 0, y = 16. x > 3 S y = -5 

12. x = 0, y > 17. x = 2, y < 1 

13. x = 0, y < 18. xy = 

19. If it = {-3, 7), and 

(a) if S is the point where the vertical line through H intersects the 

x-axis. what is the abscissa of S? The ordinate of S? What are the 

coordinates of S? 
(I) if T is the point whore ihc horizontal line through /{ intersects the 

y-axis, what is the abscissa of 7? The ordinate of T? What are the 

coordinates of 2*? 



11.3 Graphs In a Plane 447 

20. Of the following points, 8nd three that are eollinear: (3, -5), (5* 7), 
(—5, —5), .-5, 2), (ff, —5)* 

21. Describe the set of all points in the xj/*plaiie for which the abscissa is 
—2; for which the ordinate is 6. Describe the intersection of these two 
set\ 

22. Describe the set of all points in the xi/plane for which the abscissa is 
zero; the ordinate is zero, Describe the intersection of these two sets. 
Describe the union of these two sets. 



11.3 GRAPHS IN A PLANE 

A graph is a set of points. To draw a graph or to plot a graph is to 
draw a picture that suggests which points belong to the graph. The 
picture of a graph shows the axes, but they are not usually a part of the 
graph. Of course, a subset of the axes is often a part of a graph. 

It is customary to label the x- and y-axes as shown in Figure 1 1-5. 
It is usually desirable to label at least one point (other tlian the origin) 
on the x-axis with its x-coordinate and at least one point (other Ulan the 
origin) on the y-axis with its (/-coordinate. 







* 






4 
3 
2 

: 


■ 


















































* fc 




-4-3- 


1 2 3 4 






















_1 


















Li 































































Figure 11-5 

In setting up an xy-coordinate system we start with three distinct 

points O, /, / such that 57 X 0?and Ot =_0/ = 1. as in Figure 11-1. 
Although it is understood that Ol and OJ arc congruent segments 
(based on lengths in the distance system that we consider to be fixed), 
it is sometimes helpful, particularly in applied problems, to take points 
T and / so that IT! and OJ "appear" to be different in length. If this is 
done, a picture of the xt/-plane may be described by saying that the 
"scale" on the x-axis is different from the "scale" on the y-axis. The 
word scale, as used here, is not part of gilt formal geometry. As far as 
our formal geometry is concerned the distance from O to I is the same 
as the distance from O to / regardless of appearance. 



448 Coordinates in a Plane 



Chapter 1 1 



If the scale on the x-axis is different from the scale on the y-axis, a 
graph may appeal 1 distorted. An answer to the question "When is a 
square not a square?" might be "When different scales for the & and 
y-axes are used in graphing its length and width/' For example, Figure 
il-6 shows a picture of a quadrilateral ABCD all of whose angles are 
right angles. Since AB = CD = 4 and AD a BC = 4 in our formal 
geometry, it is true that ABCD is a square in our formal geometry. In 
physical (Informal) geometry, if we were to measure the sides of quad- 
rilateral ABCD with a ruler, wo would find that Ihey are of unequal 
length and conclude diat ABCD is not a square . 



Figure 11-6 





t 


*v 




1 


— 


12 


WIO 




cm 10) 


■ 


r 


T 


1 






4 


Ai2 t 6) 




fltf.G) 






















^ 




11 


12 3 4 5 6 








1 ! 11 





If a graph contains only a few points, it may be desirable to write 
the coordinates of each point beside the dot that represents iL 

Example 1 Plot the points A(-4, 3), B(0, -5), C{5, -2), D(4, 0). 
(Sec Figure 11-7,) 

















*y 






1 7]" 


























































*i 


4,3 


) 
























































































0(4,0) 




*l 








-3 




Li 


3 


















I'll 


-2) 




















-,5 




































9 


%-fy 












































































.- 







Hxwe 11-7 

Sometimes a small open circle is used to indicate that the endpoint 
of a segment or of a ray does not belong to the graph. In this connec- 
tion recall the symbol for a halflme introduced in Chapter 2. 



11.3 Graphs in a Plane 449 
Example 2 Draw die graph of {x t y) ; x < 2 and y = 3}. (Sec Figure 









4 


'V 




! 






























































2 




































x k. 






-; 






J 






3 




r 



































































































Figure 11-S 

If there are infinitely many points in the graph, the picture may 
contain segments or curves, and sometimes shaded regions or arrows, 
to indicate which points belong to the graph. 

Example 3 Draw the graph of {fr y) : 1 < x < 3 or 2 < y < 4}. 
(See Figure 11-9.) 



>5 


■IE IE 


H t 






f 

Urn , fc u 


«.- 


L : 


«-' — - 

if 


— — " "* m 2 

1 




i 


-5 


I * 









-% " 




, . 






i 1 1 





Figure 11-9 

Note that in Figure 11-9 we have shown l± and h as solid lines to 
indicate that they are a part of the graph. We have shown parts of mi 
and J7»2 as dashed lines to indicate that these parts do not belong to the 
graph. Of course, segment AD on mj and segment BC on m 2 are part 
of the graph. It is desirable to indicate all of lines vi] and m% in some 
manner since they are a part of the "Iwundary" of the graph. Let us 
agree, then, that if a line, a segment, or a ray is not part of a graph, but 
serves as a boundary to a graph , it will be shown in the graph as a 
dashed line, segment, or ray. 

Write the coordinates of the point of intersection of lines 1% and m-i 
in Figure 11-9. Is this point a point of the graph? 



450 Coordinates in a Plane 

Example 4 Daw the 



Chapter 11 
of {{x, |f) :*< -2}. (See Figure 11-10.) 




Figure 11-10 



As Figure 11-10 suggests, the graph of {(*, y) : x < -2} is the 
half plane on the left of the vertical line /. The interiors of the rays 
shown in the graph are intended to indicate this halfplane. Why does 
line I appear as a dashed line in this graph? Since line I is a vertical line, 
it is parallel to the y-axis. Therefore all the points of line I are on the 
same side of the y-axis in the xy-plane. Another way of describing the 
graph in Figure 11-10, then, is as the halfplane with edge / on the op- 
posite side of line I from the y-axfc. Since line / represents the set of all 
points hi the ,ri/-plane whose abscissa is — 2, another way of describing 
line / is the line with an equation x = — 2, The halfplane pictured in 
Figure 11-10 is the set of all points (x, y) such that x < — 2; or we may 
describe it as all of the *r/-plane which lies to the left of the tine 
x= -2. 



EXERCISES 11.3 

1. If A = (-2, 2), B = (3, 3), C = (4, -2), D = (-3, -3), draw the 
graph of the set of all points winch belong to the polygon A BCD. 

2. Draw the graph of the set of points that belong to the interior of the 
polygon in Exercise 1 . 

3. In your graph for Exercise 2, should the segments 7iB, T5C, CD, DA~ ap- 
pear as dashed lines or as solid lines? 

4. Draw the graph of 

{(x, y) : x = 3 and — 1 < y < 2}. 

5. Is the point (3. 2) part of the graph for Exercise 4? Is the point (3, — 1) 
part of the graph? 



11,3 Graphs in * Plan* 451 



hi Kxercises 13, trvsph lllfi led of all pimiK ;>', (/) in un Xgf pkne satisfying 
die given conditions. Then describe the graph of the set in words. Exercise 
6 has been worked as a sample. 

6. 2 < x < fi and y = 3. 

Graph: 



' 


* 




















































it&3j 5(6,3) 










2 








































x t 


-2 


2 








n 
























































































1 







Figure 11-11 



7. 



10. 
1L 
12. 
13. 
R 



15. 

16. 
17. 
18. 



20. 



Dcscriptiou: The graph is the segment whose endpotnts are 

A{2, 3) and B(0, 3). 
x > -2 and y = 2 
x = 5 and i/ > 
*>3 

K = -3 and -2 < r/ < 5 
—5 <x< — lor2<i/<5 
— 5 < x < — 1 and 2 < y < 5 

Which, if any, of the following points ure not part of the graph for Ex- 
ercise 12: ( — 5. 2), ( — 1, 2), ( — 1, 5), ( — 5, 5)? Which of these points are 
not part of the graph for Exercise 13? 

If P = (2. 0) and Q = (9. 0), what is the length of segment ¥Q? Justify 
your conclusion. 

If A = (2, 5) and B a (8, 5), what is the length of ATJ? Why? 

If C = { -3, - 4) and D = ( -3, 6). what is the length of CD? Why? 

Give the ctHirdinates of the midpoints of the segments whose endpoints 

are the following: 

(a) (3, 2), (3. 12) 

(b)(M>,(9.4) 

(c) (-3,1), (7,1) 

chau-kn'CE probi-em. Give the coordinates of the midpoint of the 

segment whose endpoints are (3 T 5) and (8, 7). 

challenge pkoblem. If F = (5, 1) and Q = (8, 6), what is the exact 

length of FQ? 



452 



Coordinates m a Plane 



Chapter 1 1 



11.4 DISTANCE FORMULAS 

Consider an ^-coordinate system as shown in Figure 1 J -12. Let J 
be the unit point on the it-axis and let P and Q be any two points with 
al jseissas % l and x 2 , respectively, on the .t-axis. 









'*>y 








































































































i £ 9 x 







«1 3 x 2 























































r^giire IMS 

We know, by the definition of a coordinate system on OX relative to 
unit segment Ol (Definition 3.3), that 

PQ (in 01 unite) = s, - x 2 \ = (** - «,|. 

For example, if P = (3, 0) and Q = {7, 0% then %i = 3, x* = 7, and 

PQ = |3 - 7| = |7 - 3J = 4. 

Now suppose that I is any line in the xy-planc and parallel to the 
.T-ajds as in Figure 11-13 (that is, I is any horizontal line), Let P and Q 

be any two points on I and let Pi(x u 0) and Qi{x 2 , 0) be the projections 
of P and Q t respectively, on the x-axis. 



Q 



3— t* 



f, 



0: 



■i": 



i 2 



Figure 11-13 

By our definition of the abscissa of a point in an ry-plane, we know that 
the abscissa of Pisxj and that the abscissa of Q is x 2t If P = Pi and 

Q = Q u then 

PQ = PyQi = fa - X2J. 



11,4 W>tar»ce Formula* 453 

If P=£P lf then Q=£Qi and the quadrilateral PPiQi() is a parallelo- 
gram. Therefore we again have 

PQ = P t Q, = \x, - x 2 \. 

Note that iflisa horizontal line, then / is perpendicular to the y-axis 
(Why?) and every point of I projects onto the same point in the i/-axis. 
Thus all points on a horizontal line have the same ordinate. We have 

proved the following theorem. 

TJJEOBE \f 11.2 If p( Xit y T ) and Q(x 2 * ij\) are points on the same 
horizontal line in an xi^plane, men 

PQ = |*i - *!. 

It should he noted that if P = Q in Theorem 11.2. then 
x, a Xt and PQ = = 0. 

TllEOR EM 11.3 UP(x l ,y 1 ) and Q(x if f/ 2 ) are points on the same 

vertical line in an .viy-plaiie, then 

pq - m - y«l- 

Proof: The proof is similar to the one given for Theorem 11.2 and is 

assigned as an exercise. 

Example I If A = (3, 5) and B = (-6, 5), find AB. 
Solution! AB = [3 - (-6)| = 9, 

Note that if two points have the same ordinate, then they lie on the 

same horizontal line. Thus A 6 in Example 1 is a horizontal line. Hence 
AB is die absolute value of the difference of the abscissas of the points 
A and B t as shown in the solution to Example 1. 

Example 2 If C = (-2, -3} and D = (-2, 4), find CD. 

Solution: CD m -3 -4 = |-7| = 7. 

Note that if two points have the same abscissa, then they lie on the 

same vertical line. Thus CD in Example 2 is a vertical line. Hence CD 
is the absolute value of the difference of the ordf nates of the points C 
and D, as shown in the solution to Example 2. 



4W Coordinates in a PJarre 



Chapter 1 1 



You have seen how to find the distance between any two points in 
an ay-coordinate system if the two points lie on the same horizontal line 
or if the two points lie on the same vertical line. Next we show how to 
find the distance between two points if Lhey lie on the same oblique 
line, that is, a line that is neither horizontal nor vertical Before pro- 
ceeding to the general case, let us consider an example. 

Exampk3 UP = (-2, -3} and Q = (4, 5), find FQ. 

Solution: {See Figure ! t - 14.) Let h be the line through Q and parallel 
to the y-axis and let h be the line through F and parallel to the x-axis. 
These twojines intersect in a point R such that h ± h at R, Why? 
Therefore FQ is the hypotenuse of a right triangle, &PQR, with the 
right angle at R, Since f 2 is a horizontal line and l± is a vertical line, we 
have 



PR = |4 - {-2}| = 8 
QR = |5 - (-3)1 = 8 

by Theorems 11.2 and 11.3. 
By the Pythagorean Theorem, 

to 2 = m? + (<?w, 

= 62 + 82, 
= 36 + 64, 
= 100, 
and hence 

PQ = 10, 











,55 




t*i 














i 














A mm 














/ i 










| / 




/ 


' 


' \ 


-A 


! * 






i 


4- 


~H 




_ ld__ k 




-2,-3) 




]R<A,-3) ~^ 






1 1 


I 


i 



Figure 11-14 



Note that in finding PQ in Example 3 we made reference to a right 
triangle. We proceed now to the theorem that, will enable us to find the 
distance between any two points in any ^-coordinate system without 
reference to a right triangle. The formula in this theorem is often re- 
ferred to as the Distance Formula for point 1 ; in an xu-plane. 

THEOREM HA If F, = (*,, pj and P 2 = (x 2t y 2 ) are any two 
points in an xy-plane s then 

PiPi = V(*T- *a) a + (m - y 2 ) 2 . 

Proof: There are four cases to consider. 

Cam L Fi = F 2 . 

Case 2. F L and F 3 are distinct points on a horizontal line. 

Case 3. Fi and P 2 are distinct points on a vertical line. 

Case 4. Pi and F 2 are distinct points on an oblique line. 



11,4 Distance Formuttt 455 



Proof of Case 1: If Pi = P** then x± = xj, yi = yz y and, by the for- 
mula of Theorem 11.4, we have 

P& = Vo = o, 

as it should since the distance between a point and itself is defined to 
be zero. 

Proof of Case 2: If P t and ?2 axe distinct points on the same hori- 
zontal line, dien y± = tj2 and, by the formula of Theorem 1 1 .4, we have 

PlP2 = \Z(X! - X 2 )2 = h - X 2 . 

This result agrees with the statement of Theorem 11.2 for two points 
on the same horizontal line. 

Proof of Case 3: If Pi and P 2 are distinct points on the same vertical 
line, then Xi = x% and, by the formula of Theorem 11.4, we have 



PiC«i.ja 



P1P2 = y/(y L - 1/2) 2 = |yi - U2I 

which agrees with the statement of Theorem 11,3. 

Proof of Case 4: If P t and P 2 are distinct points on an oblique line as 
shown in Figure 11-15, then the line through Pi and parallel to the 
y-2Lsis intersects the line through P% and parallel to the x-axis in a point 
H(xy, r/2) such that APiP^H is a right triangle, 

P 2 R = [xi - %z\ 
by Theorem 112 and 

Piil = Ij/j. - y a \ 
by Theorem 1 1 .3. We have 

(p 2 Rr- = ,*, - x 2 p = (x, - 

and 

Since PiPz is the hypotenuse of AP1P2R, we have, by the Pytha- 
gorean Theorem, that 




J^*2.j'a) 



or that 



PxP* ~ V 
and the proof is complete. 



(P1P2) 2 = (P 2 H) 2 + (PxH) £ 

fi - X2P + (#, - ya) 5 . 



456 Coordinates in a Plane Chapter 11 

Since (xi - $$* = (x 2 — *i) 2 and (y t - y 2 ) 2 m (y 2 - ifi) 2 . in 
using the Distance Formula of Theorem 11.4, it does not matter which 
point is designated P t (x lt tji) and which point is designated Pg(* 2 » y 2 )> 

Example -1 If A = (2, -4) and B = (-5, 3), find AB> 

Solution: Substituting the coordinates of the given points in the Dis- 
tance Formula, we have 

AB=W- (S)p + (-4 - 3)4 

= yya + (_ 7) 2 

= y/49 + 49 
1 



= 7\/2 

Note that in working Example 4 we considered A(2, — 4) M the 
point J*i(xi, 1/1) and B( — 5, 3) as the point Pafet Jte) when we substi- 
tuted these coordinates into the distance formula. Show, hv working 
Example 4 again, that we would have obtained the same result for AB 
had we considered A(2, — 4) as the point Pz(x 2 , y 2 ) and B( — 5, 3) as the 
point Pxfa, ^i). 

EXERCISES 11.4 

1. Prove Theorem 1].3. 

In Exercises 2-10, use Theorem 1 1 .2, Theorem 1 1 ,3, or the distance formula 
to find the distance between the given points. 

2. (-3, 2) and (-3, 11) 

3. (f I© and <& ID 

4. (-2.5, V^ and (17.3, y5) 
3. [w, 4.8} and fa -9.6) 

6. (-2, 7) and (3, -5) 

7. (3. -6) and (9, 0) 

8. (4 r 17) and (-3,9) 

9. (-3, -5) and (5, -1) 
10, (6, -3) and (-4, 2) 

IL Find the perimeter of die triangle whose vertices arc A( — 2, —3), 
B(3, 9), and C(- 10,12). 



1L4 Distance Formula* 457 

12. Prove that the triangle whose vertices Are P{1> 2), Q(9 t 2), and /l(5, 8) 
is isosceles. 

13. AAHC lias vertices A (6. 0), B(-4 t 4), and C(10, 4). 

(a) Find the perimeter of A ABC. 

(b) Find the area of AABC. 

14. Find the lengths of the diagonals of a quadrilateral ABCD if 
A = (4, -3), B = (7, 10), C = (-8, 2), and D = (-1, -5}. 

15. The vertices of APQR are P(- 1, -2), ()(4, 0), and fl(2, 5). Prove that 
A PQR is a right isosceles triangle. 

16* If the distance between A(B t -2) and B{0 t tj) is 10, find the possible 

tf^oordtnates of B* 
17* Find the coordinates of the points on the x-axis whose distance from 

(2, 8) is 10. 

IS. If (a, — a) is a point in quadrant IV, prove that the triangle with vertices 
(—5. 0) f (0, 5), and (o T — a) is isosceles. 

19. Given D = (-2, 2), E = (10, 2}, F = (4, y) with Z DKE a right angle, 
find the two distinct possible values of jf. 

20. Given P = (-2, -7), Q = (3, 3), R = (6, 9), use the distance formula 
to show that PQ + QR = PR, and hence that F-Q-R. 

■ Exercises 21-26 refer to the triangle whose vertices are A = (2, 4), B = 
(6, 8), and C = (12, 2). 

21. Draw line /i through A and parallel to the r/-axis. Draw line / 2 through 
B, parallel to the x-axis, and intersecting f t at D. What are the coordi- 
nates of D? 

22. Draw line h through C, parallel to the x-axis, and intersecting h at E. 
What arc the coordinates of £? 

23. What kind of quadrilateral is quadrilateral BDEC? Find \BDEC |. 

24. What kind of triangles are ABDA and ACEA? Find \ABDA\ and 

AGFA . 

25. Copy and complete: 

BDEQ m \AABC\ + \A[T\\ + \&JB , 

26. Find \AABC\. 

27. challenge problem. Find \APQR if P = (6, 0), Q = (1, 5), and 
R = (10, 8). 

2& CHAU.EXCE problem. If A = (0, 0), D = (b t c), B = {«, 0), and C = 

(a ■+■ b* c), where a, b t c are positive numbers, prove that quadrilateral 
ABCD is a parallelogram. (Hint If the opposite sides of a quadrilateral 
are congruent, then the quadrilateral is a parallelogram.) 



4M 



Coordinates in a Plane 



Chapter 11 



11.5 THE MIDPOINT FORMULA 



We know that the midpoint M of a segment PQ is the point be- 
tween P and Q such that PM = MQ or thai PM = JPQ, or, similarly, 
M@ = ^P(X Suppose that the coordinates of P and Q arc given and 
that we wish to find the coordinates of M, the midpoint of J 5 ^. For ex- 
ample, if P = (2, 3) and Q = (8, 5), we can find the coordinates (# t/) 
of M, the midpoint of FQ t as follows. (See Figure 11-16.) 



'■' 


1 








aotn 


t 


M"i0.y) 






Mix: 






■-<* 


£r 














l«*3> 






1 


















! 








i 






















i i 




X 







^(2,0) M'{x,0) Q'£8,0) 










| 1 


















! ! 







Ffcm Hie 

Let F(2 T 0), AT(x, 0) t (X(8, 0) be the projections of P, M, Q, re- 
spectively, on the *-axis and F'(0, 3 )* W'{0, i/), Q"(0, 5) be the projec- 
tions of P, M, Q f respectively, on the y-axis. Since If is the midpoint 
of FQ, it follows from Theore m i 10. S that M' is the midpoint of FQ' 
and M" is the midpoint of Fp 77 . Therefore, by Definition 3,3, 
2 < x < 8 and 3 < t/ < 5< By the definition of midpoint, we have 

(1) FW = M'Q' and (2) V'M" s M'V ■ 

By Theorem 11.2, 

FM' = \x - 2| and Afp* = |8 - *|- 

Sinee x — 2 > (Why?) and 8 - a: > 0, we get by substitution into (1), 

x - 2 = 8 - x. 

Therefore 2x = 10 and x = 5. 

Similarly, by Theorem 11.3, 

F'M" = \y-3\ and M"Q' f = \5 - y\. 

Since y — 3 > and 5 — y > 0, we get by substitution into (2), 

y - 3 = 5 - y t 

Therefore 2y = 8 and y = 4 Hence the coordinates of .W, the mid- 
point of PQ, are (5, 4), 



11.5 Th« Midpoint Formula 459 



We can proceed, as in the preceding example, to find the coordi- 
nates of the midpoint of any segment if the coordinates of the end- 
points of the segment arcs given. However, Theorem .1 1.5 provides us 
with a formula that will enable us to find the coordinates of the mid- 
point of a segment. The formula in Theorem 1 1.5 is often referred to as 
file midpoint formula. 

THFnREM 1L5 UP = {x ly yi) and Q = {x 2 , y 2 ) are any two dis- 
tinct points in an xy -plane, then the midpoint M of FQ is the point 



»= fr^> 



There are three cases to consider. 

Cose 1: P and Q are distinct points on a horizontal line. 

Case 2: P and Q are distinct points on a vertical line. 

Case 3: P and Q are distinct points on an oblique line. 

We shall begin the proof for Case 3 and assign the remainder and 
the proofs of Cases 1 and 2 as exercises, 

Procifof Case 3: Let P(xi,y t ) and ()(X2, 1/2) be any two distinct points 
on an oblique line in an sy-plane as shown in Figure 11-17 and let 
M(x, y) be the midpoint of FQ. Let F(xi, 0), M'(x, 0), and Q'{x 2 , ()) be 
the projections of P> M, an d Q, r espectively, on the %-axis. By Theorem 
10.3, M' is the midpoint of r<^ r . By Definition 3.3, x is between xi and 
X% dius 



<*< x% 



or 



X'l < * < Xt< 



(In Figure 11 -17 we have shown x± < x < x 2 , but this order might be 
reversed if P and Q are chosen in a different way.) 






<jfa2,^) 




F(*1,Q) M'lx,0) Q-Cx 2l CI) 



Figure IM7 



460 Coordinates in a Plane Chapter II 

By the definition of midpoint, 

par = arg\ 

and by Theorem 115, 

FAT = \x - %i\ 



and 

M'Q f = ,** - 4 
Therefore 

\X - Xi\ = |-t2 - X\> 

If *i < i < x 2 , then x — xi > and *a — x > and we have 
x — *i = x 2 — x. Why? Therefore 

2x = *t 4- x 3 
and 

" = —2— 

If ^2 < x < xi, then xi — x > and x — x 2 > 0. In this case, 

FM' = |x - ri. = xi — x 

MV = |x 2 - x\ = x - z 2 . 

x — x 2 = Xi — x, 
and again we have 

-_ * + » 

2 ' 

We have shown that if M (x, y) is the midpoint of the segment whose 
endpoints are P{x yy iji) and Q{x 2 , ys), then the abscissa of \f is * 2 . 

In a similar way, it can be shown that the ordinate of M is ** 1 _ ^ 2 , 

(Yon arc asked to show this in the Exercises, thus completing the proof 
of Case 3.) 

hxamplc 1 Find the coordinates of the midpoint of the segment 
whose endpoints are the following: 

1. A(2. -3) and 5(2, 9). 

2. C(-12, 1) and D(-3, 1). 

3. E(-2, 7) and F{W t 12), 



11,5 Th* Midpoint Formula 461 

Solution: 

L Segment XB lies on a vertical line. Therefore i = x\ = *t and 

*i + *£ *i + JCi 

—IT = ~2— "* 

Therefore the midpoint is 

2. Segment CD lies on a horizontal line. Tlierefore y = t/i = 1/2 

tfi + ya _ yi ■+■ tji __ 

2 "-* 1 ' 





2 




Therefore th« 


j midpoint is 


Ma 


V 2 * 


yi) 




/-12 + (- 


-3) 



.»)-(-*»> 



3. Segment EPlies on an oblique line; hence the midpoint M of EF 

is the point 

Example 2 The vertices of a triangle are A(0, 5), B(4 f 3), and 
Q-2, 1). Find the length of the median to BC. 

Solution; The median to BU is the segment whose endpoints are 
A{Q t 5) and M, the midpoint of BC, The coordinates of M are (1, 2) by 

the midpoint formula; hence 

AM = y/(Q - If + f£ - 2) 6 

= VTT9 
= V10 

by the Distance Formula. 



462 Coordinates in a Plane Chapter 1 1 



EXERCISES 11.5 

1 . Complete the proof of Case 3 of Theorem 11.5 by showing that the ordi- 
nate of M is Hl±Jil , (See Figure 11-17.) 

IS 

2. Prove Case I of Theorem 11.5. (In this case, y = tj\ = tj2 in the state- 
ment of the theorem.) 

3» Prove Case 2 of Theorem 1 1 .5. (In this case, x = i] = r* in the state- 
ment of the theorem.) 

■ In Exercises 4-10, find the midpoint of AB if A and B have the given 

coordinates, 

4. (-5, -2) and (-5, 6) 

5. (-3, 5) and (8,5) 

6. (0, 0) and (8, 10} 

7. (0,0) and (-8, 10) 

8. {1, 2) and (6, 14) 

9. (r, 7) and (3r, -3) 

10. fab)md(-5a,7b) 

11. The vertices of a triangle are I\2, -3), £{10, 1), and K(4 t 6). Find the 
midpoint of FQ. Find the length of the median to FQ. 

12. The vertices of a triangle are A(-3, -2), B(l, 6), and C(5, -2), Find 
the lengths of the medians lo A~B and BC« How are the lengths of these 
two medians related to each other? What kind of triangle is A ABC? 

■ Exercises 13-15 refer to quadrilateral ABCD whose vertices are the points 
A = (0, 0), B = (6, 0), C = {8, 4), D = (2, 4). 

13. Find AC and BIX 

14. Show that the midpoint of AC is the same point as the midpoint of B7X 

15. What kind of quadrilateral is ABCD? 

M E*ercises 16-24 refer to A ABC whose vertices are the points A = (2, 0), 
B = (12, 0), C = (7, 5\/3). 

I K. Prove that A ABC is equilateral. 

17. Find the coordinates of D, the midpoint of AB. 

18. Find the coordinates of E, the midpoint of fflOL 

19. Find the coordinates of F, the midpoint of AC. 

20. Show that the lengths oi' the three medians of A ABC are equal. 

21. Find DE and show that t>E = JAC. 



11.5 The Midpoint Formula 463 

22. Find EF and show that EF = JAB. 

23. Find FD and show that FD = \BC. 

24. Do the results of Exercises 21-23 prove that ADEF is equilateral? 

Exercises 25-27 refer to A/SK whose vertices are the points / = (0, 0\ 
S = (6, 0), K = (0 

25. What kind of triangle is A JSK? 

26. Find the coordinates of M, the midp>int of 5K. 

27. Prove that JM s ]$K, 

In Exercises 28-31, the coordinates of two points A, M are given. Find the 
coordinates of the point B suck that M is the midpoint of 3uB, 

28. A = (1,3),M={4,7) 

29. A = (4, 7), Af = (L 3) 

30. A = (-1, -8), M = (0, 0) „ 

31. A = (-6, -4), M = (-3, -2} 

Exercises 32-36 refer to the segment AH whose endpoints arc A = (3, 2), 
B= (11, 6) and the point Pi x> y) on AB such that AP = %AB. 

32. Let A', P\ /T be the projections of A, J* B. respectively, on the .t-axis. 

33. What is the absciss u of F in Exercise 32? What is the abscissa of P? 

34. Let A", P", B" be the projections of A , P, B t respectively, on the (/-axis. 

35. What is the ordinate of P" in Exercise 34? What is the ordinate of P? 
Mi, What are the coordinated of PI 

37. If A = (-4, -2) and B = (6, 3), find the coordinates of the point P on 
AB such that AF = JAR (ERnfc See Exercises 32-36.) 

38. challenge problem. Given positive numbers a and h and a right tri- 
angle with vertices at A = (0, b) t B s (a, 0), and C = (0, 0), find the 
coordinates of Af, the midpoint of A~B t and show that CM = ^AB. Does 
tills prove that for antj right triangle, the median to the hypotenuse is 
one-half as long as the hypotenuse? 

39. challenge PROBLEM. Given positive numbers a, h, and u quadri- 
lateral ABCD with vertices at A = (0, 0), B = (a, 0), C = (a + b, c), 
and D = (b, c), 

(a) prove that ABCD is a parallelogram, 

(b) prove that the diago nals AlT and ED bisect each other by showing 
that the midpoint of AC is the same point as the midpoint of B~D. 



464 Coordinates in a Plane 



Chapter 11 



11.6 PARAMETRIC LINEAR EQUATIONS 

If we are given a line in an xjy-ptune, it is often desirable to find the 
coordinates of points on that line. We know that two distinct points 
determine a line. Suppose that we know the xi/-coordinates of two dis- 
tinct points A and B on AB. We should be able to find the xiy-coordi- 
nates of any other point P on AB provided, of eoursc 5 enough informa- 
tion is given to determine one and only one point P on AB. 

Example I Let A = (1, 3), B = (4, 7), and suppose that S, Q, fl are 

^ -y v ^ fr 

points on AB such that 5 is on AB, Q is on AB, and R is on opp AB, 
[See Figure 11-18.) Suppose, further, that 

AS = $AB, AQ = 2AB, AB. = AB, 

and we wish to find the coordinates of S, Q t and iL 



4 


iy 




J* 












J 








7$ 










/ 


. 






/ 1 1 








*£:- 






6 






Ml 






X s 




































;i 




/ 


B) 






















flfa 




































































X 






~ 3 di 




b 




















/ ~ 3 















Figure 11-18 

Let S = (x, y). To find x and y we might proceed as follows. Let A\ 
B', S f be the projections of A, B % S, respectively, on the r-axis, and let 
A", B" t S" be the projections of A, B, S, respectively, on the y-axis. 
Using Theorem 10.8, we get 



2 _ 

3 " 


AS _ A'S' *— 1 
AB " A'B' " 3 


2 
3 " 


AS A"$" y - 3 
AB A"B" 4 



and 



and 



.r = 3, 



9 3 



11.6 Parametric Linear Equations 465 

Similarly, we can find that Q = (7, 11) and R - (-2, - 1). (Verify 
these results by selling up appropriate equations and solving Ihcm.) If 

Fwere any other point on AB such thai 

AP=kAB, 

where k is any positive number or zero, we could find the coordinates 
of P by a computation similar to those above. Our objective, however, 
is to derive an expression from which the coordinates of any point on 

AB can be obtained by simple replacements. 

In Chapter 3 we studied a coordinate system on a line. In this chap- 
ter we have already defined an xi/-coordinate system in a plane, based 
on two line coordinate systems like those you studied in Chapter 3, 

Let us consider now a third line coordinate system, a coordinate 
« — * 
system on line AB with A as origin and B as unit point We call it the 

^-coordinate system on AB, It should be clear that we start with an 
,tf/-coordinate system based on a unit segment for distance. When we 
speak of "the distance*' between two points in die .ry-plane, we are 
talking about the distance based on the same unit segment that is used 
in setting up the x^-coordinate system. For every different choice of 
points A and Bona line fin the i^-planc, there is a different k coordi- 
nate system on I with A as origin and B us unit point. The ^-distance 
between two points on / will usually be different from the ^(/-distance 
between those points. 

The following table shows the ^-coordinate, the {/-coordinate, and 
the ^-coordinate of the points A, B, S, Q, B that were shown in Exam- 
ple L 



Point 


x-Coordinate 


y-Coordinate 


fe-Coordiiiatc 


A 


1 


3 





B 


4 


7 


1 


S 


3 


s§ 


I 


Q 


7 


11 


2 


R 


-2 


-1 


-1 


P 


m 


13 


k 



We shall derive the equations which show us how to compute the 

x- and {/-coordinates of a point on AB in terms of its fc- coordinate. First 
we shall show that the ^-coordinates of the points of I form a coordinate 
system on I as do the ^-coordinates. 



.:i,i. 



Coordinate* in a Plane 



Chapter 11 



Let / be any nonvertical line in an .vy-plaue. (See Figure 11-19.) Let 
O(0, y ), fft yi), P(x, y) be points on l y and let O t l\ F be their re- 
speetive projections on the x~axis. It foflows from the Plane Separation 
Postulate and the properties of parallel lines that (J r t t F have the 
same betweenness relation as do their respective projections. For ex- 
ample, if O-I-F. then O-l'-F. 




Figure IMS 

Let § denote the unique coordinate system on I with O as origin 
and I as unit point. Note that Ul serves as a unit segment for § and that 
generally (except if / is parallel to the Y-axis) distances "in g" arc dif- 
ferent from distances "in the xtf-coordinate system.*' Let p be the co- 
ordinate of F in the system §, Since betweenness foT points on I is the 
same as for their projections on the x-axis, it follows from the definition 
of a coordinate system on a line that p and x are both positive, or both 
negative, or both zero- It follows from Theorem 10.8 that 



OP 



O'F 

or ' 



Since these ratios are equal regardless of the distance function, we have 



OP 
OI 



;i _ 1 ir " or "1-0 



= Jx|, and p as x. 



Now S is the one-to-one correspondence that matches each point P on 
/ with a number p. Since p = x, we see that the correspondence that 
matches each point of I with its x-coordinate is indeed a coordinate 
system. Similady if lis a non horizontal line, then the correspondence 
that matches each point of I with its y-ccKM-dinate is a coordinate system 
on I. We state these results in our next theorern. 



TlfEQREM 11.6 If I is any nonvertical (nonhori/ontal) line in 
an xiy-planc, then the one-to-one correspondence between the 
points of / and their x-coordinates ((/-coordinates) is a coordinate 
system on I. 



11.6 Parametric Linear Equations 467 

In the next theorem, as well as in many others throughout the rest 

of this book, there is the assertion, 'A is real" within a set-builder sym- 
bol. Ttiis is short for "k is a real number/' 

THEOREM IL7 If A(x ls yi ) and B(x 2t y 2 ) are any two distinct 
points, then 



AB = {(x y) : x = x, + % 2 - XjJ.y = {/! + %a - J/i),fcisreal}, 

If fc is a real number and if P is the point (x t y) where 
x = *i + fc(x 2 - «!) and ^ = yi + k(y 2 - fiX then 



fe a 



AT 



and 



P€ A§ 



if 



fc> 0; 



"*-4l and P£oppA$ if Ar < 0. 

JVoo/: Let Aixi, y t ) t B{x 2 , f&) be any two distinct points. Suppose first 

that AB is neither vertical nor horizontal, Think of three coordinate 

systems on AB as suggested in Figure 1 1-20, the x-eoordinate system 
and the y-coordinate system, determined by the xy-coordinatc system 
(see Theorem 11.6), and the ^-coordinate system in which A is the 
origin and B is the unit point. 




Figure 11*20 



or 
or 



468 Coordinates in a Plane Chopttr 11 

It follows from tlic Two-Coordinate-Systems Theorem (Theorem 

3.6) and its corollary, each applied twice, that AB is the set of all points 
P(x 3 r/) such that 

x — xi k — y — iji _ k — 

x-2 — xx 1 — * y2 — yi 1 — * 

x - xi = *(* 2 - Xi\ y - i/! = % 2 - yt); 

x = X! + *<*2 - *i). y = yi + %2 - yi); 

where ^j^. 

Since the definition of a coordinate system requires that hetween- 
ness for points agree with betweenness for coordinates, it follows that 

P is on a5 if and only if h ;> 0, and ? is on opp AB if and only if k < 0, 

If A E is a vertical line, then the x-coordinate of every point P on AB 

is die same number, and there is no x-eoordinate system on Ad, In this 

case xi = xo = x for every point P{x t y) on AB and the equation 

x = *! + *(* 2 - *i)* 

which simplifies to i = ii, is still applicable. The proof for the noii- 
vertieal-nonhorizontal case is applicable to this case as far as the rela- 
tion between y and k is concerned. Therefore the assertion of the the- 
orem applies in the vertical case. Similarly, the theorem may be proved 
for the horizontal case. 
Tn 

S3 = {(x, y) : x = x t + k(x 2 - %§}, y = y± + k(y 2 - yi), k is real}, 

the "it is real" is put in because k is not mentioned before the braces 
and not before the colon within the braces, and we read the sentence as 

"AB is the set of all points (x t y) such that 

x = X! + k(x 2 - x a ), 

y = yi + %2 - yi), 

where k is real/' 
The equations 

X = x l + k(x-i - Xi) and y = i/i + %z - f/i) 

hi the statement of Theorem 11.7 are called parametric equations for 
die line AB and k is called the parameter. A parameter j s usually 



11.6 Parametric LI near Equations 469 

thought of as a variable to which values may be assigned arbitrarily, 
and other variables are defined in terms of it. By assigning real number 

values to the parameter k in the parametric equations for AT?, we ob- 
tain ordered pairs of numbers corresponding to points on AB in the xy- 
eoordinate system. Of course, we could assign values to x (in the non- 
vcrtieal case) and find values of y so that the resulting ordered pairs 

correspond to points of AB or we could assign values to y (in the non- 
horizontal case) and End values of x so that the resulting ordered pairs 

correspond to points of AB. But it is easier, and it works in all cases, to 

assign values to k and compute values of x and of y so dial the resulting 

ordered pairs correspond to points of AB. 

Note that parametric equations for a line aie not unique. If / is a 
line, then there are many choices for the two points A and B of The- 
orem 11.7, and hence many pairs of parametric equations for L 

If we replace the coordinates (% x 3 ) and (yj, j/g) in the parametric 
equations of Theorem 1 1 .7 with the coordinates (1, 3) and (4, 7) of the 
points A and B in Example 1, we obtain the parametric equations 

x = 1 + 3k and y = 3 -I- 4k 

for the line AB in that example. Recall that, in Example 1, S is a point 

on AB such that AS = |AB. Thus, by replacing k with the number f 
In the parametric equations 

X m 1 + m, y = 3 + 4Jt, 

we obtain the coordinates (3, 5j) of the point S. Show that you get the 

xy-coordinates of the points Q and R on X5 in Example 1 by assigning 
the values 2 and — 1, respectively, to k in the parametric equations 

x = 1 + 3fc, y sb 3 + 4k. 

It follows from Theorem 1 1,7 that every line can be represented by 
set-builder notation and a pair of parametric equations. However, it is 
not true that every pair of paTametric equations represents a line. For 
example, the set 

S = {(*, y) : x = 2 4- k ■ 0, y = 3 + k ■ 0, k is real} 

is a set whose only clement is (2, 3). Note that in the statement of The- 
orem 11.7, the points A and B are distinct; hence not both of the co- 
efficients of k in the parametric equations can be zero. That is, we can 
have x 2 — xi=0 or y2 — tji=0, but we cannot have both x$— xj = Q 
and y2 — yi = 0. 



470 Coordinates In a Plane Chapter 1 1 

Although not eveiy pair of parametric equations represents a line , 
our next theorem provides us with a method for identifying those pairs 
of parametric equations that represent lines. 

THEOREM 11.8 If a t h, c, d are real numbers, if b and d are not 
both zero, and if 

S = {(*, y) : x = n + M, y = c + dk, k is real}, 
then S is a line. 

Proof; Taking k = and k = 1, vvc get two points in S, namely 
A (a, c) and B(a + b t c + rf )• Tty Theorem 11.7, parametric equations 

for AB are 

sc = fl-r*ft(a-Hfc — a) = a + M 

and 

tfs«-rx(6+a-c)s@-fJ& 

Therefore A3 = {(x f y) ; x = a + bkj y = c -\- dk,kis real}; 

hence S = An and S is a line. 

hi addition to being able to write parametric equations for a line, 
we can also write parametric equations for a segment or a ray if we 
place the proper restrictions on the parameter k. We illustrate this 
technique in the following examples. 

Example 2 Let A = (4, 1) and B = (2, —3). Using coordinates and 
parametric equations, express (1) AB, (2) AB. and (3) AB. 

Solution; 

1. Substituting the coordinates of A and B into the equations of 
Theorem 11,7, we get x = 4 - 2k and y = 1 - 4k as para- 
metric equations for AB. Therefore 

AB ' = {(x, y) : x = 4 - 2k, y = I - 4k t k is real}. 

2. If P is a point on AJ?, then P £ A? if and only if Jt > 0, Therefore 

AB = {(*. y) : * = 4 - 2fe y = 1 - 4^ * > 0}, 

3. If P, A, B are three points on AB and if &i, & 2 , *a are the coordi- 
nates of P, A, /I, respectively, then Pis between A and # if and 
only if fci is between feg and fife. Since the ^-coordinates of A and 

B on AB are and 1 , respectively, then P is between A and B if 
and only if k is between and 1. Therefore 

m = {(x t y) : x ^ 4 - 2k >y = I - 4^0 < k<, } }. 



11.6 Parametric Lln«ar Equations 471 

Example 3 Given A = (4, 1} and B = (2, -3), find P on AB such 
that AP = 4 'AB, 

Solution: Taking k = 4 in the parametric equations of Example 2, 

we get 

P = fry) = (-4, -15). 

Example 4 Given A = (4, 1 ) and B = (2 t - 3), find P on opp AB such 
AP = 4 • AB, 



Solution: Taking k = — 4 in the parametric equations of Example 2, 
we get 

P = fr y) = (12, 17). 

Example 5 Given A = (4, I) and B = (2, -3), find C and D % the 
points of bisection of AB. 

Solution: Taking fe = -j- and ft = -J in die parametric equations of 
Example 2, we get C = (J£ , — -§-) and D a (|, -}) as the points of 
bisection. 



EXERCISES 1LG 

In Exercises 1-5, the coordinates of two points A, B are given. Use para- 
metric equations and set-buiJdcr notation to express AB t AB, opp AH, and 

m 

L A = (1, 4), B = (3, 7) 4, A = (2, 3), B = (0, 0) 

2. A = (2. 2). B = (5, 5) 5. A = (-3, -2), 3 = (0, 1) 

3. A =(-1,3), B = (3,0) 

G-IO* Find the coordinates of the midpoint of AB in Exercises 1-5. {Hint: 
Let k = -j in the parametric equations for AB.) 

1 1. Find the points of bisection of AB in Exercise 2. 

hi Exercises 12-17, a relation between AP and AB is given, If 
A = (-2, -5) and B = (4. 1), find the coordinates of P in Z8. 

12* AP = 2AB 15, AP = t^2AB 

13. AP = 25AB 10. AP = LSAff 

14, 4AP = MB 17. AP = \AB 



472 Coordinates in a Plane Chapter 1 1 

18-23. The instructions for Exercises 18-23 arc the same as for Exercises 

12-17, except that F is in opp AB, Kecall that -j^ = —kin this case. 

In Exercise 18. AP = 2AB as in Exercise 12; in Exercise 19, 
AP = 25AB as in Exercise 13; etc. 



24. Find the coordinates of P £ AB if A = (-1. 5), B = (4, -3), and 
AP = 3A#. (There are two possible answers.) 

25. Given A = (0, 4), B = (3„ 0), and (7 is on AB, find the ^-coordinate of 
C if its x-coordinate is — 2. (Hint: Obtain the parametric equations for 

AB and let x = —2 in one of these equations (Which one?). Solve this 
equation for k and use this value of k to find the (/-coordinate of C.) 

2a Given A = ( - 1, 3), B = (2, -3). and C is on A$ t 

(a) find the {/-coordinate of C if its x-coordinatc is 5. 

(b) find the abscissa of C if its ordinate is 8. 

In Exercises 27-31, draw the graph of the set. 

27. {(*, ij)\\+3k,y = 2- k, k is real} 

28. {(x, y}:x = 3k.y = k,0<k<3} 

29. {(x, y):x= -2 + k,y = -2k, k > 0} 

30. {(*, y) : x = -K y = K * <, 0} 

31. {(x, y) : x = 5, y = 2 + Jfc, -2 < k < 3) 

Exercises 32-39 refer to the triangle whose vertices are A = (3, 2). 

B = (9, 4), and C = (5, o), 

32. Kind the coordinates of D, the midpoint of AB. 

33. Find the coordinates of E, the midpoint of BC, 

34. Find the coordinates of t\ the midpoint of AT. 

35. Find the coordinates of the point R in CzJauch that CR s 5CD. (Use 

the parametric equations for CIJ, where C =■ (xi, yi) = (5, 8). 
D = (x 2 , i/ 2 ) = (6, 3), and A- = f ) 

36. Find the coordinates of the point S in BF such that #S = JBF. (Use the 
parametric equalions for BF, where B = (xt, t/i), F = (x 2 , 1/2), and 

37. Find the coordinates of the point Tin AE such that AT = JAE. 

38. Is Jt = S = 77 

39. Show Uiat tlie three medians of A ABC intersect in a point whose dis- 
tance from each vertex is two-thirds of the length of the median from 
that vertex. 

40. challenge problem. If a, b, e are positive numbers and if A = (0, 0), 
B == (a, 0),. and C = (b, c), show that the three medians of A ABC are 
concurrent at a point whose distance from each vertex is two- thirds of 
the length of the median from that vertex. 



11.7 Slope 473 



11.7 SLOPE 

In this section we develop the idea of the slope of a line. Slope 
corresponds to the idea of the steepness of an inclined plane or the 
steepness of a stairway. If all the steps of a stairway are uniform, we 
may describe the steepness of the stairway in terms of the "rise" and 
"run** of one of its steps. The steepness of a uniform stairway is the 
number obtained by dividing the rise by the run of one of the steps. 
For example, we may say that the steepness of the stairway shown in 
Figure 11-21 is |. 




Figure U-Sl 

In our formal geometry, we define the steepness or slope of a line 
in a similar way. That is, we define the slope of a line in terms of the 
slope of any segment on the line and we define the slope of a segment 
in an xy-plane in terms of the coordinates of its endpointa. In informal 
geometry, we can think of the slope of a segment in somewhat the 
same way as we think of the slope of a step in a stairway; that is, in 
terms uf its "rise divided hymn." Figure I 1-22 suggests that the slope 
of AJ9 is $ = -J-. Note that in terms of the coordinates (2, 3) and (8, 7) 
of the endpoints A and B of AB, the rise is |7 — 3| = 4 and the run 
is |8 - 2| = a 



Figure 1 1-22 







kj 










- 1 1" 




















V*M 






H 
















Ri 














j* 




90 


















I 






3 










.- 1 . 






A(*3) 




Run C{8, 3) 












































v k 











9 


































Coordinates in a Plane 

Thus, if A = (x±, 1/1) and B 
the slope m of A3 as 



Chapter 11 

= (x^y j/y), where *i =?^ *2» we c.otdd define 



m = 



x a — Kl 



hut we do not. The same formula without the absolute value symbols 
is not only easier to handle but it is more useful Hie sign of the slope 
Indicates whether the segment "slopes up or down." In Figure 11-22, 
segment AB has a positive number for its slope and we note that the 
segment slopes up as we view it from left to Tight. However, the seg- 
ment CD shown in Figure 1 1-23 slopes down as we view it from left to 
right 





- 


'■• 


111 I 














ioMB 














TCI 








— 


a 






1 \ 












! \ 












BWU \ 


D(7 r 1) 






J l 




B 


1 | * ■ 







Figure 1143 

If we use the formula 

x 2 - Xl 
in compute tlie slope of CD, we obtain 



1-5 

7-4 



4 

3' 



a negative number. This is one of the reasons why we do not include 
the absolute value symbols in our formula for the slope of a segment. 
That is, if the slope of a segment is a positive number, the segment 
slopes up when viewed from left to right, and if the slope is a negative 
number, the segment slopes down when viewed from left to right. We 
are ready now for our formal definition. 



Definition 11.2 If A(xi, y%) and B(% j/a) are two distinct 

points and if Xx =£ x*, then the slope of AB is * /g "~ * 1 . 

X-i — Xi 



If A(xu tfi) and B{x% t tjz) are two distinct points and X\ = #2, then 
"KB is a vertical segment and slope is not defined for 'KB in this case. 



11.7 Siope 475 



If A(*i, i/i) and B(x2, 1/2) are two distinct points and r/j = y 2 * then^B 
is a horizontal segment and its slope is zero. 

Note that in computing the slope of a segment A B it docs not mat- 
ter wiiich endpoint of AB is designated (afe 1/]) and which endpoint is 
designated fa, 1/2)- Thus, for the slope m of AB in Figure 11-22, we 
could write 

m = y*_zJ± = LzJL = 1 = 1 

' x 2 - xi 8 - 2 "" 6 "" 3 
or we could write 

__ j/g - yi 3-7 _ _-4 _ 2^ 

X2 - Xi " 2 - 8 " -6 ~ 3 * 

Similarly, for the slope m of CD in Figure 1 1-23, wc could write 

Xz — Xx 7-4 3 

or 

Xb — xi 4—7 3 

As suggested earlier, the concept of the slope of a line is based on 
the concept of the slope of a segment. It seems reasonable that the 
slopes of all segments of a nonvertical line are equal, and we state this 
as our next theorem. 

THEOREM LL9 The slopes of all segments of a nonvertical line 
are equal. 

Proof: Suppose that AB is any nonvertical line with A = (x u y t ) and 

B = (x$, ys). Then, by Theorem 11.7, AB can be expressed parametri- 

cally as 



= {(x, y) : x = xi + k{x- 2 - x T )» ij s t^ -|- % 2 - ^i), fc is r 
By Definition 1 1 ,2, the slope of AB is 

X 2 — XL 

Let R and S be any two distinct points of AB. There are two distinct 
numbers k] and h$ such that 

R = (x-i + fei{xa - *i), i/! + Ai(y 2 - J/i)) 
and 



476 Coordinates in a Plane Chapter 1 1 

By Definition 11.5, the slope of ESh 

fa + k 2 (y 2 - iji)) - (y-i + Ky 2 - j/i)) _ (</a - t/i)fe - fr) 
(*"i + k 2 ix 2 — xij) — (x\ + k\(x2 — *i)) {x 2 - xt)(ki - ki) 

X2 - «J 

We have proved that any two -segments of AB have the same slope, and 
!hr pn lof is complete. 

Definition 11.3 The slope of a oonvcrtical line is equal to 
the dope of any of its segments. The slope of a non vertical 
rav is equal to the slope of the line that contains the ray. 

Example 1 Find the slope of the line which contains the points 
A = (-2,5) and B = (4,0). 

Solution: AB is a nonvertieaJ line and, by Definition 11.3, the slope 
of AB is equal to the slope of A~B. Therefore the slope of AH is 

y 2 - *Ji _ 0-5 _ _£ 
*2-*i "4-<-2)~ 6' 

We know that two distinct points determine a line in the sense that 
if any two distinct points are given, then there is exactly one line which 
contains them. It is also true that a non vertical line is determined by 
any point on it and its slope. That is, there is exactly one line which 
contains a given point and has a given slope. We now state tins 
formally, 

THEOREM 11.10 Given a point A and a real number m » there is 
one and only one line which contains A and has slope m. 



Existence. Let a point A = (*i, yi) and a slope m be given. Let 
B be the point (x\ + 1, «/i 4- m). Then AH is a line which contains A 
and has slope m. (Show that the slope of AB is m.) 

Uniqueness. Let PQ be any line which contains A and has slope 

m. Since FQ is a nonvertical line (Why?), it intersects the vertical line 
through B(xi + I, y 1 -f m) in some point R(xi — 1, y 2 ) as suggested 
in Figure 11-24. (The figure shows R and B as distinct points. We shall 

prove that they are actually the same point.) The slope of FQ is equal to 



11.7 Slope 477 




Figure U-S4 

Therefore y 2 = jft + «b Hence H = B and Py sa AB, This proves 

that A/5 is the only line through A with slope m, and the proof is 
complete, 

If wc know the coordinates of two points on a line, we can use The- 
orem 11.7 to write parametric equations for the line. Theorem 11.10 
implies that a line is determined by any point on it and its slope. There- 
fore we should be able to write parametric equations for a line passing 
through a given point and having a given slope. Our next theorem tells 
us how to do this. 



THEOREM I hi I The line I given by 

1. I s {(*, y) : x = *i + fc, y = y, -f mk, k is real} 
or by 

2. I = {{x, y) : x = xi + sk t y = y t + rfc, k is real) 

is the line through (x lt y\) with slope m = — . 

Fnx>f: We shall establish (J) and (2) separately. 

Troof of 1: Taking k = and k = I in the parametric equations in 
(1), we get (x% t i/i) and (*j -f 1, y± + m), two points on I. The slope of 
the segment joining these two points is 

yi + m - yi _rn_ m 
*i + 1 - xi 1 

Therefore I as given in (1) is the line through (%, i/i) with slope m. 



478 Coordinates in a Plane Chapter 11 

Proof of 2: Taking k = and k a i in the parametric equations in 
(2), we get (*i, yi} and (x t + $, y^ + r), two points on I The slope of 
the segment joining these two points is 

j/i + r - yi _ r 
*i + & - Xy $ 

Therefore I as given in (2) is the line through {x^ t/i) with slope rn, and 
the proof is complete. 

Example 2 A line p passes through (2, 5) and has slope 3. Find the 
point on p whose abscissa is — 2. 

Solutimt: Using Theorem 11,11, we can express p as 

p = {(x, y) : x = 2 + ft, y = 5 + 3fc, k is real). 
We next set % = -2, obtaining -2 = 2-|-&,orfc= -4. Hence 

y = 5 + 3(-4) = -7. 
The point on p whose abscissa is —2 is (— 2, —7). 

Example 3 Find the slope of the line 

I = {(x, y) : x = 2 - 3fc # = 3, Jfc is real}. 

SoJi*rio»: Taking k = and k = 1, wc get (2, 3) and ( — 1, 3), two 
points on the line. The slope of the segment joining these two points is 

3 - 3 _ 



-1-2 -3 

Therefore the slope of f is zero. Is f a vertical, horizontal, or oblique 
line? 

We conclude this section with the following summary remarks re- 
garding the slope of a line. 

L If a line has a positive number for its slope, then the line is ob- 
lique and slopes up when viewed from left to right, (See Figure 
ll-25a.J 

2. If a line has a negative number for its slope,, then the line is ob- 
lique, and slopes down when viewed from left to right, (See Fig- 
ure ll-25b.) 

3. If a line has slope aero, then the line is horizontal. (See Figure 
ll-25c.) 

4. Slope is not defined for a vertical line. 



11.7 Slope 479 





Zero 



■lOM 



(C) 
Figure 11- 



EXKRCI5ES 11.7 



In Exercises 1-10, find the slope of the segment joining the given |>oints. 
Express each answer as a fraction in lowest terms. 

1. (2, 5) and (5, 7) & (2^ -4) and (-1, 4\) 

2. {- 1. -3) and (8, 3) 7. (-5.6, 3) and (1.4, -£) 

3. (0, -4) and (8, 0) 8. {1.2. -5) and (3.2, -5) 

4. (- 1, 4) and (2, -2) 9. (2, 5) and (5, 2) 

5. (-3, -3) and (3, 3) 10. (7. 12) and (1, -3) 

11^20. In Exercises 11-20, tell without plotting the points whether the 

given line (a) slopes up, (b) slopes down, or (c) is horizontal. In Kxercisc 
1 1, the line is the one that passes through the two given points in Ex- 
ercise 1. In Exercise 12, die line is the one that passes through the two 
given points in Exercise 2, and so on. 

21. (a) On the same xt/-plane. graph the line r through the two points of 

Exercise 1 and the line s through the two points of Exercise 2, 

(b) What appears to be true about these two lines, that is, how are r 
and ^ related? 

(c) How are the slopes of these two lines related? 

22. (a) On the same xy-plane, graph the line p through the two points of 

Exercise 3 and the line q through the two points of Exercise 4. 

(b) What appears to be true about these two lines, that is, how are p 
and q related? 

(c) How arc the slopes of these two lines related? [Hint: Find the prod- 
uct of the slope of p and the slope of q.) 

In Exercises 23-27, find x or y (whichever is not given) so that the line 
through the two points will have the given slope, 

23. (4, 1) and (x, 4), m = 3 26. (4, -3) and (0, y), m =s -| 

24. (-3, 1) and (6 t y) t m s -\ 27. (5, 12) and ( -2, y), m - 

25. (x, 0) and (-3, 5), m = J 



480 



Coordinates in a Plana 



Chapter 11 



28. (ti) On the same xtf-plane. graph the line I through the two points of 

Exercise 25 and the line n through the two points of Exercise 2ft 

(b) What appears to be true about these two lines, that is, how are J and 
n related? 

(c) How are the slopes of I aud n related? 

29. (a) Plot the quadrilateral ABCD with vertices A = (— 2, —3), 

B = (3, -2), C = (4, 2), and D = (~1. 1). 

(b) Which pairs of sides have the same slope? 

(c) Is A BCD a parallelogram? 

30. Write parametric equations for the line I through { — 2, 5) with slope 2. 

31. Write parametric equations for the line p through (1, - 1) with slope -J. 

32. Use the parametric equations for line / obtained in Exercise 30 to find 
the coordinates of at least one more point on I and plot the graph of 1. 

33. Use the parametric equations for the line p obtained in Exercise 31 to 
find the coordinates of at least one more point on p and plot the graph 
of p. 

34. One way to plot the graph of line p in Exercise 31 is suggested by the 
figure. Note thai the point Q{1, - 1) is on p and the "slope fraction" f 
tells us how to get from one point to another on p. The numerator 3 
represents the difference in the ordinates of two points on p, and the de- 
nominator 2 represents the difference in the abscissas of the same two 
points. Thus, if we begin at the point (1, —1). which is known to be 
on p, we arrive at another point on p by increasing the ordinate and 
abscissa of [L — 1) by 3 and 2, respectively. If the slope is negative, it 
can be written as a fraction with a negative numerator aud a positive 

denominator; for example, —^—, In tliis example we could get from one 

point to another, in informal geometry, by moving 2 to the right and 3 
down; in formal geometry, by adding 2 to the abscissa and subtracting 
3 from the ordinate. Write the coordinates of the points A T B T C. 





. 


":• 








fp 
















% / 




















/ 




















: ; i 
















s 








2 
































a 






















1 


/ 
J 






















\ 














a 


/ x % 









m- 


■ 








1) 






































/I 





















11.8 Other Equations of Lint* 4-8 1 

35* Use the method described m Exercise 34 to plot the graph of the line 
through (—2, 1) with slope y. 

36, Use the method described in Exercise 34 to plot the graph of the line 
through (0, 6) with slope -J, 

37, Use the method described in Exercise 34 to plot the graph of the line 
through (0. 0) with slope 2. (ffinf; 2 = f ) 



11.8 OTHER EQUATIONS OF LINES 

In Section 11.6 we showed how a line can be expressed in terms of 
parametric equations. In this section we shall show how parametric 

equations for a line can lie used to obtain other equations of the line. 
You are already familial" with most of these equations from your work 
in algebra. Perhaps the simplest equations for lines are those for hori- 
zontal and vertical lines. 

THEOREM 1L12 If Ms the horizontal line through {x lf 1/1), then 

'= tfey) '■ y = »)■ 

Proof: Let (jcfc, 1/2) he any other point on the horizontal line I through 

the point [t\, iji). By Theorem 11.7, 

1 = (fr y) : * = *i + H*2 - *i), y = yi + %s - yi)> k k real}. 
Tine I is horizontal so tj\ = y 2 - Therefore 

y = yi + %2 - #1) = !/i + *-0 = y T . 

For any real number fc» 

* s x* + k(x2 — x%) 

is a real number. Conversely, for every real number ,t» there is a real 
number k such that x = Xt + fc(*2 — x i)- Th^ one an ^ on ly ^ that 
works here is 



*2 - ^l 



It follows that I is the set of off points (at, «/) such that j/ = r/ t and x is a 
real number, that is, that 

This completes the proof. 



462 Coordinates In a Plane Chapter 1 1 

Iii the statement of Theorem 11.12 it should be clear that we could 
say that 

1 = {(*> y) - x m Xi + k{x 2 - Xi), y = i fl , k is real}, 
or 

I = {(x t if) : x is real and ij = yi} t 
or 

' = {(*; ST : y = §Ji. * is real}* 

Since the equation x = xi + A(* a — *i) establishes a onc-tc-otie cor- 
respondence between the set of all real numbers thought of as k- values 
and the set of all real numbers thought of as x-v&lucs. these set-builder 
symbols all denote the same set. 

THEORTM 1L13 If I is the vertical line through {x u ft), then 
/ = {(x, y) ; x- *i}. 

Proof: Assigned as an exercise. 

Example 1 If P = { — 4, 3), write an equation of (1} the vertical line 
through P and (2) the horizontal line through P. 

Solution* 

1. An equation of the vertical line through Pis x = —A. 

2, An equation of the horizontal line through P is y = 3. 

Our next theorem tells us how to write an equation of a line if we 
know the coordinates of any two distinct points on the line. 

THEOREM 11.14 {The TwthPoint Form) Tf A = (x u y\) and 
B = (x2, 1J2) are distinct points, and if AB is an oblique line, then 



= {fr„) :£=*-= JLzIl) 

I X 2 — Xi 1/2 - Ui) 



*2 - x t y 2 - y^ 

Proof: We are given that A = {xi, tji) and B = (x z , ife) are distinct 
points on oblique line AB< There are two things to prove. 

1. If P = (x, u) is a point of Aft, then — = " "~ - 1 - . 

2 - If ^3^ = ^7=^ ^ P = ^ $ '^point of £2. 



UJB Oth« Equations of Lin** 433 

Proof of 1; Suppose that F a (x, 1/) and that P is a point on A/3, If 
P = A, then 

JL^L = . and JLZliL^XlllL. 

if* - yi *2 - *i y* - yi 

If P ^ A, then by Theorem 1 1 .9 the slope of AT equals the slope of AS 
since AP and A B are segments (not necessarily distinct) of the same line 

aS. But the slope of £B is ^ ~ * jl and the slope of A? is ^~ & . 

x 2 — Xi x — Xj 

Therefore 

ya-¥i _ g-ffi 

X 2 - X, "" X - Xi " 

Multiplying both sides of this last equation by x — Xi, then dividing 
both sides by y 2 — y& then simplifying, we get 

* - *i _ y - yi 

*2 — %i 1/2 - yi " 
Therefore, in all cases, whether A = P or A=?=P t 'd P £ AB, then 

* - *i _ y - tfi 
ac 2 - xi y a - 1/1 

Proof of 2: Suppose that P = (x, 1/} and that 

x-xi _ y- yi 
Xi - x, " y 2 - yi " 

Either * — Xi = or x — x± ^= 0. If x — xi = 0, then 1/ — j/ a = 0, 

(a, y) = (xi, yi), P = A, and P € AB, If x — x, ^£ 0, multiply both sides 
of 

*-*t _ y-yi b y 2 - yi 
*2 - xi Jfa — yi x - xi 

to get 

Uz-lfi y-Ui 

Xg — Xi X — Xl 

Since ~ 5EL is the slope of AB and *• *± is the sloi>e of AP, this 

#2 — *l X — Xi 

proves that if x — x x =j£ 0, then the slopes of AB and AT 5 arc the same 
and, as we shall prove, P Q AB. 



484 Coordinates in a Plane 



Chapter 11 



Suppose, contrary to what wc assert, that the slopes of A H u i id . \ / ' 
are equal, but that A } B, P are noncollinear as in Figure I. .1.-26. 




Figure 11-26 

Let Qfe y') be the point in which the vertical line through P 

intersects AB. It follows from Theorem 11.9 that the slopes of A%) and 
AB are equal, so the slope of A$ is equal to the slope of AP, and 

y* -y\ _ y -yi 

X — Xt X — JTi * 

y' - ui = y - y^ 
ft = y> 



and the points A, B, P are collinear. Since our supposition that A, B, P 
are noncollinear leads to a contradiction, this proves that our suppo- 
sition is false. Therefore A, S, P are collinear and P lies on AB. This 
completes the proof in all cases, whether x — Xi =: or x — X\ y^ 0, 

that if 

* - *i _ y - yi 
*2 - *i "" y* - yi 

and P = fc y) t then F £ AB. This completes the proof of Theorem 
11.14. 



The equation 



* - *i _ y 



yi 



yi- yi 

in the statement of Theorem 1 1 . 14 is often written in the form 

(i) y - yi = *±=3Ht - x,). 

*2 — 3.1 



11.8 Other Equations of Lines 485 



Show how Equation (1) is obtained from the one in the statement of 
Theorem 11.14. Since A = (acj, y\) and fi = {x^ y%l are distinct points 

on the oblique tine AB of Theorem 11.14, then 



-2i = m 



— *1 



is the slope of AB. Substituting m for ^ ^ in Equation (1), we 



(2) rj - y t = m(x - xx) 

for AB. It should be noted that Equation (2) still holds if XS is a hori- 
zontal line since, in this case, m = and Equation (2) reduces to 
y = yi. Thus, if we know the slope of a line and the coordinates of any 
point on the line, we can write an equation of the line. When an equa- 
tion of a line is written in the form of Equation (2), it is often referred 
to as the point slope form. We have proved the following theorem, 

THEOREM 1L15 (The Point Slope Form) If Hs the line through 
A = (xu yi) with slope m, then 

'={(*> y) ' v - *j\ - »*(* - *i)} 

Example 1 If A a ( — 2, 3) and B = (4, ft), write an equation of 

Solution: Substituting the coordinates of A and B in the Two-Point 
Form of an equation, we obtain 

* - (-2) y-3 
4 - (-2) 6-3 
or 

x 4 L 2 _ y -3 
6 ~ 3 

as an equation of AB. 

IF we multiply both sides of the equation in Example 1 by 6, and 
add — 2y 4- 6 to both sides, we obtain 

x - 2y + 8 = t 

which is of the form 

Ax + By + C = 0, 

with A = 1, B = -2, and C = 8. 



4-86 Coordinates "in a Plane Chaptarll 

This latter form is often referred to as the general form of a linear equa- 
tion, that is, of an equation whose graph is a line. Although we shall not 
do so here, it can be proved using the theorems of this chapter that if 
A, £, C are real numbers with A and B not both zero, then Ax + By + 
C = Ois an equation of a line. The converse statement is also true, that 
is, every line has an equation of the form Ar + By + C = in which 
A, B, C are real and A and B arc not both zero. What is the graph of 
Ax + By + C = if A a 0, B = 0, C=£0?ifA = 0\ B = 0\C = 0? 

Example 2 Write an equation of the line through (3, —4) with slope 
— ■§ and put the equation in general form. 

Solution: Substituting (3, — 4) for {%%, grj and — -| for m in the Point 
Slope Form of an equation, we obtain 

i, + 4 = -|<r - 3). 

Multiplying both sides of this last equation by 3 and adding 2* — 6 to 
both sides, we obtain 2i + 3i/4-6 = 0asan equation of the line in 
general linear form. 

Every nonvertfcal line intersects the y-soas (Why?) in a point 
whose coordinates are (0, h), where b is a real number. The number b 
is often called the y -intercept of the line. Similarly, every nonhorizontal 
line intersects the z-axis in a point whose coordinates are (a, 0), where 
a is a real number. The number a is called the x-mtercept of the line. It 
should be clear that die x- and {/-intercepts of a line can be obtained 
from an equation of the line as follows: 

1. If the line is a nonhorizontal line, the ^-intercept is obtained by 
substituting for tj in an equation of the line and solving the 
resulting equation for x, 

2. If the line is a tionverrical line, the (/-intercept is obtained by 
substituting for x in an equation of the line and solving the 
resulting equation for y. 

Example 3 Find the x- and t/-tntcrccpts of the line whose equation is 

3s - 4y + 8 = 0. 

Solution: Substituting for y i u the equation 3x — Ay -f 8 = 0, wc 
obtain 3x + 8 = 0, or x = — 2$. Hence the ^-intercept is — 2|. Sub- 
stituting for x in the equation, we obtain — Ay + 8 = 0, or y = 2. 
Hence the {/-intercept is 2. 



If we substitute (0 T b) for {%i, ij\ 
equation of a line, we obtain 



in 



11.8 Other Equations of Unas 487 

the Point Slope Form of an 



or 



y — b = m(x — 0) 
tj = mx + b. 



It should be clear that when an equation of a nonvertical line is put in 
the form ;/ = mx + b, the slope and (/-intercept of the line can be read 
directly from the equation, This form, which is often called the slope 
y-intercepl form, is especially convenient when one wants to draw T tlie 
graph of a line whose equation is given. 

Exanipte 4 An equation of the line / is 3* + 2y — 8 = 0. 

L Put the equation in slope (/-intercept form. 
2. Draw the graph of I on an xy-p\m\c. 

Solution: 

1. To put the equation 3x + % — 8 = in slope (/-intercept form 
we add — 3x 4- 8 to both sides and then divide both sides by 2. 
The resulting equation is u = — Jar + 4. 

2. The graph of t is shown in Figure 1 1-27. Note that the slope and 
(/-intercept of I can be read directly from the equation y — 
— Jx -(- 4, Thus, if we start at the point where y = 4 on the 
y-axis [that is, the point (0, 4}] and "trace out" a slope of — §, 
we arrive at a second point on /, These two points determine 
/ and henee they determine the graph of I. 









I 


- 


V> 


















































».4)1 


5 2 






















C~ i 
















- 3 






V 










\ X 


-5 









JV_ 5 





















































Figure 11-27 

As you might expect, two nonvertical lines are parallel to each other 
if and only if they have the same slope. If you worked Exercises 22 and 
28 of Section 11.7, you should have discovered a relationship between 
tlie slopes of two lines that are oblique and perpendicular. We state 
these properties of parallel nonvertical lines and of perpendicular 
oblique lines as the last two theorems of this section. 



488 Coordinates in a Plane 



Chapter 11 



THEORESf 11.16 Two non vertical lines are parallel if and only 
if their slopes are equal. 

Proof; Let two non vertical lines r and 9 be given. We have two things 
to prove. 

L If r || $, then their slopes are equal. 

2. If the slopes off and s arc equal, then r s. 

Proof of 1: Suppose that r and * are nonvertieal lines and that r [f $, If 
r = s T then r and s are the same line and hence they have the same 
dope. Suppose r^=s t as shown in Figure J J -28. Let l\xu yi) and 
Q(xt, 1/2) be two distinct points on r. Let the vertical lines through 
P and Q intersect s in F[xi, iji + k) and Q'{x 2> j/2 + k), respectively. 




Figure 11-2H 



Then PFQ'Q is a parallelogram (Why?) and PF = Q^. But PF = \h\ 
and QQ 1 = k\. Why? Therefore \h\ = \k\. Since h and h are either both 
positive or both negative, wc have h = L The slope of Pfp (and hence 



of r) is 



*2 — *i 



.The slope of FQ' (and hence of s) is 

(jfe + A) - (yi 4- h) _ y 2 -y 1 
%2 — *! X2 - Xj ' 



since ft = fe Therefore the slopes of r and s are equal. 

Proo/ o/2: Suppose that r and s are nonvertieal lines and that their 
slopes are equal. If r = s ( then by definition r I .9. Suppose that r =£ s. 
Let m be the slope of r. Then s also has slope m. Now, either r and s arc 



11.8 Other Equations of Lines 489 

parallel Of they are not. Suppose they are not parallel. Then they have 
exactly one point P(a"i, r/i) in common. Tin is we have two distinct lines 
passing through the same point and having the same slope. This con- 
tradicts Theorem 11.10, Therefore our supposition that r and -s are not 
parallel is incorrect; hence it follows that r || $ and the proof is 
complete. 

THEOREM Il.il Two oblique lines are perpendicular if and 
only if the product of their slopes Is — 1. 

Proof. Let lj and Z 2 be two given oblique lines with slopes mi and wig, 

respectively. We have two statements to prove. 

1. If /i _L h, then mi • mg = — 1 . 

2, Umi*m 2 = — 1 , then h _ k- 

Before proceeding with the proof we wish to comment on the 
adjective "oblique" in the statement of the theorem. Would the state- 
ment that results if "oblique" is erased be a theorem? No, it would not. 
For if one of two lines is not oblique, then those two lines arc perpen- 
dicular if and only if one of them is horizontal and the other is vertical. 
Since a vertical line has no slope, there would be no product of slopes 
in this case. 

We now proceed to the proof of statements 1 and 2. 

Let p\ and p% be the lines through (0, 0) and parallel to h and fa, 
respectively, as shown in Figure 11-29. Then the slope of pj is mi and 
tie slope of p% is nio. Why? Since neither »i nor »2 is a vertical line, 
they both intersect the line 

I = (fc tj) : x = 1}, 



Figure 1 1-23 p\ 




490 Coordinates in a Plane 



Chapter 11 



Let A = (1, gft) and B = (1, 1/2) be the points of intersection of I with 

lines pi and p 2 , respectively, as shown in Figure ] 1-30, 




Figure 11-30 

Tli en the slope of p\ is 



and the slope of p 2 is 



Wi -0 



y 2 -0 
m 2 = *— ~ = y 2 . 



Therefore A = (1, mi) and B = (I, ma). Now, h _L l 2 if and only if 
pi 1 pz- 

It follows from the Pythagorean Theorem and its converse (applied 

to AOAB in Figure 11-30) that 

Pl _ p 2 if and only if (OAf + (OB)* = (AB) 2 . 

From the distance formula, we get 

(QA)* = 1 4- wn 2 , 
(OB) 2 s 1 + mtf* 



and 
Thus 



(AB) a = (ms - mi)*. 

if and only if 1 + mi 2 ■+■ 1. 4- m 2 2 ss (mg — mi) 3 , 

if and only if 2 -j- w-i 8 + «»^ = m 2 2 — 2m im^ + »»i a , 

if and only if 2 = — 2f9t3!fft& 

if and only if mim? = — 1, 

Therefore £1 _L Z 2 if and only if m jm 2 = — 1, and the proof is complete. 



11.8 Other Equations of Lines 491 
Example 5 If 

h =((*,!,): 5*- 2y + 4=0}, 

h = ((jtv y) : 2a: + 5y - 15 = 0}, 

and 

fe ■ {(a& </) : 5* - 2# - 8 = 0}, 
show that l\ I ?3 and that ti ± h. 

Solution; Putting the equations of I it f 2 , /g in slope ^-intercept form, 
we get 

h - {{*> V) : </ = $* + 2}» 
I 2 = {(*,</): y = — $* + 3}> 
and 

k = {(J, y) : y = fx - 4}. 

Since ^ = slope of li = slope of h> it follows that 1$ \\ I3, Also, since the 

product of the slopes of l-\ and 1% equals 

*•(-« = -i. 

It follows that /i X fe, K »s also true that (3 J- k- Why? 

EXERCISES 11.8 

1. Prove Theorem 11.13. 

In Exercises 2-6, use the Two-Point Form to write an equation of the line 
containing the given points, and put each equation in general form (that is, 
the form Ax + By + C = 0). 

2. (1, 5) and (3, 4) 5, (<X 0) and ( - 1, 6) 

3. (0. 3) and (-5, 0) 6. (-2, 2) and (2, -2) 

4. (0 t -3) and (5, 0) 

In Exorcises 7-12, use the Point Slope Form to write an equation of the line 
which contains the given point and has the given slope, and put each equa- 
tion in general form. 

7. (3, 5) and m = 1 

8. (-2, l)andm= -1 

9. (a0)andm = | 

10. (5, 0) and m = -| 

11. (-3, -7)audm = -1 

12. (-5, -3)andm = | 



492 Coordinates in a Plane Chapter 1 1 

■ In Exercises 13-16, determine which word, parallel or perpendicular, would 

make a true statement. 

13. The lilies of Exercises 7 and 8 are |T] . 

14 The lines of Exercises 8 and 1 1 lire [Tj and distinct. 

15, The lines of Exercises 9 and 12 are [TJ . Are these two lines distinct? 

16. The lines of Exercises 9 and 10 are [?]- 

17* Write an equation (in general form} of the line which contains the point 
{3, 8) and is parallel to the line whose equation is 2x — 3y = 10. (Hint: 
What is the slope of the line whose equation is 2* — 3y = 10?) 

18. Write an equation (in general form) of the line which contains the point 
;3, Si and is perpendicular to the line whose equation is 2x — Zlj = 10. 

19. Write an equation (in general form) of the line which contains the origin 
and is perpendicular to the line whose equation is y = x. 

■ Iu Exercises 20-25, an equation of a line is given. In each exercise, (a) put 
the equation in slope (/-intercept form, (b) find the slope of the* line, and 
(c) find the x- and ^-intercepts of the line. 

20. 2x - 3y - 12 = 

21. 3x + 2y = 16 

2& 4x - 6y = -8 
23* y + x = 

24. 4x - 2ij = 1 1 

25. 5x + 4y + 13 m 

26. Which pairs, if any, of the lines of Exercises 20-2-5 are parallel? Which 
pairs, if any are perpendicular? 

27. Find, without graphing, the coordinates of the point of intersection of 
the lines 

P = {{*> a) : * + % = 6} 
and 

q={(x,y):5x + 4tf= -3}. 

(Hint: Note that p is not parallel to q (show this) and hence the lines 
intersect in exactly one point.) l>et (x u y±) be the point of intersection 
of lines p and q. Then, since (*j, j/j) is on p, x x + 3i/i = 6 or 

(1) Ui m - j*i + 2. 

Also since (x\, y^) is on q ¥ Sxj -f 4yi = —3 or 

m sn = -^ - 1 

Apply the substitution property of equality to Equations ( 1} and (2) and 
find xi and tji. 



1U8 Othar Equations of Lines 493 

28. Show without graphing that lines 

p={(x > t,):2x-3y=i2} 

and 

q = fc y) - 4y + x m -5} 

are not parallel, and find the coordinates of their point of intersection, 
(See fjcercise 27.) 

Exercises 29-39 refer to the triangle whose vertices are A = (1, 1), 

B = {9, 3), and C = (7 t 9). In each case where an equation of a line is asked 
for, write the equation in slope (/•-intercept form. (It may help to draw the 
figure in an xy-plane and label the points as you need them.) 

29. Find the coordinates of the midpoint D of AB. 

30. Find the coordinates of the midpoint £ of BC. 

31. Find the coordinates of the midpoint F of AC* 

32. Write an equation of the line through A and & 

33. Write an equation of the line through B and F. 

34. Write an equation of the line through C and D, 

35. Show that the lines A£. Bb\ CD intersect in the same point. 

36. Write an equation of the line through E and F. 

37. Write an equation of the line through A and B. 

38. Show that F? || XS and hence that bW | AB. 

39. Show that EF = \AB. 

40. If a # and h s£ 0\ an equation of the form J- -f- = 1 is called the 

ah 

interrupt form of an equation of a line. Show that the line whose equa- 

X ft 

tion is r i — l contains the points (a, 0) and (0, b). 

41. Put the equation 3* -f Ay = 12 in intercept form (see Exercise 40) and 
read the .v- and [/-intercepts directly from the equation. 

42. Given that p = ((x t y) : x = $}, 

(a) Is(3 5 7)€p? 
(b)ls(3,l7)ep? 

(c) b(4,4)Cp? 

(d) Is (3, -9) € p? 

43. Given that q = {(a-, y) : y = - i}: 

(a) Is(-4,5)€<j? 

(b) Is(-4, V5)€<?? 

(c) Ik(-4,-w)€tf 
(d)Is(4,-4)Cc/? 



494 



Coordinates in a Plane 



Chapter 1 1 



44. challenge problem. Use theorems of this chapter to prove the fol- 
lowing statement: If A. B, C arc real numbers with A and H not both 
zeru T then Ax + By + C = is an equation of a line. 

45. challenge problem. Use theorems of this chapter to prove the fol- 
lowing statement: Every line has an equation of the form Ax + By + 
C = in which A, B,C are real and A and B are not both zero. (Note 
that this statement is the converse of the statement in Exercise 44.) 



11.9 PROOFS USING COORDINATES 

We have defined an xy-coordJnate system in a plane and have used 
coordinates as tools in much of our work in tills chapter, Given a plane, 
there are many xy- coordinate systems in that plane. In constructing a 
proof of a geometric theorem, it is wise to select a convenient ^-coor- 
dinate system that fits the problem and, at the same time, reduces the 
number of symbols needed in the proof. Such a selection yields no loss 
of generality, yet reduces the amount and difficulty of work involved. 
We illustrate with our next theorem which appeared as a corollary in 
Chapter 10. 

THEOREM 11.18 A segment which joins the midpoints of two 
sides of a triangle is parallel to the third side and has half the length 
of the third side. 

We shall give two proofs of Theorem 11.18. In the first proof, 
we select an arbitrary x ^-coordinate system in the plane of the triangle 
without any regard to the position of the vertices and sides of the given 
triangle. In our second proof, wc "pick" an ^-coordinate system in 
the plane of the Lriangle in such a way as to reduce the number of sym- 
bols needed in the proof, 

Proof I: Let AABC in any ary-plane be given. (See Figure 11-31.) 

Cfaun) 



figure 11-31 




£(*S>«) 



11,9 Proofs (Mill Coordinates 495 

Suppose that 

A = (xi, yi) t B = (x 2j jfe), C = (x 3} y s ). 

Let D, E, F be the midpoints of BC y AC, KB, respectively. Then 
U.sinu; llic Distance Formula, we get 

(BE) 2 = (^_±ii _ *i ±j» V + / yg + ys _ g/i + ya \ 2 . 

= ft* - *i)) 2 + (Ksfc - ^)j*, 

But, by the same formula, 

{AB)* = {x 2 - x v f + (y, - m f. 
Therefore 

{DEf = i(AB)2 t 

° r DE = #AB). 

Next* we prove that DE [| KB. Suppose that AB is a vertical segment 
Then, 

xj a 3C2 Rod — - — SB — i 
Therefore 

"~l 2 ' 2~~ /* 



F _ (*i + x 3 yi + y 3 \ 



and DE is a vertical segment. Hence DE || AB. If ZB is not u vertical 

segment, then its slope is ^ 2 ~ ^ 1 . The slope of TM is 

xs — Xi 

2 2 _ y 2 - yi 

Xi + x s xi + x 3 x 2 - xi ' 



Therefore, since DE and AB have the same slope, DE [ KB. This com- 
pletes the proof so far as the segment DE is concerned. In a similar 
way, wc can_prove that EF = \BC and EF || BC and that DF = ^AC 

and DF II AC. 



496 



Coordinates in a Plane 



Chapter 11 
Proof H: Let A A B C be given . 1 n the plane of this t dangle there is an 

^/-coordinate system with the origin at A 9 with Afi as the x-axis, with 
the ^-coordinate of B positive, and with the ^-coordinate of C positive, 
(See Figure 11-32.) 



q2x&2>a) 



Figure 11-32 




B{2xi,0) 



Let x lt x 2s y-2 he real numbers such that B = (2x [3 0), C — (2x2, 2f/2}* 
Let D t £, f* be the midpoints of EC, AC, AB, respectively. Then 

D = (xi + x s , i/ 2 ), E = (x 2s y 2 )» 
the slope of DE = 0, the slope of KB = 0, 

DE a |{X! + x 2 ) - x 2 | = jxj = x lt 
and 

AJ? = 12x41 =2*|. 

Therefore DE || KB and £»E a {A#. This completes the proof for UE 
and this is all that we need to prove, since DE might be any one of the 

three segments which joins the midpoints of two sides of the given 
triangle. 

To prove the statement of the theorem for the segment which joins 
the midpoints of two sides, we first label the triangle so that AC and CB 
are those two sides and then proceed as above. Thus each of the three 
parts of the proof uses a different coordinate system, but what we write 
in each case is the same. For example,. Figure 11-33 shows another pic- 
ture of the triangle shown in Figure 11-32. However, it shows a differ- 
ent xy-coordinate system and a different labeling of the vertices. Vertex 
C in Figure 11-32 becomes vertex A in Figure 1 1-33, A becomes B t and 
B becomes C, and we have made the x-axis look horizontal. 

You should note that the proof that DE = \AB and DE \\ AB would 
proceed exactly as before, but UR in Figure 11-33 is not the same seg- 
ment DE as in Figure 11-32. The same applies for the segment AB, 



11.9 Proofs Using Coordinates **? 



C(2i 2 ,2>'2) 




Figure 11-33 



It is clear Hi at Proof II is simpler than Proof I and is to be preferred. 
In general, if a proof using coordinates involves a polygon, it is usually 
easier to construct a proof if we select an ^-coordinate system in the 
plane of the polygon satisfying one or both of the following conditions: 

1 . Let the origin be a vertex of the polygon and let the positive part 
of the x-axis contain one of the sides of the polygon, 

2. If the polygon contains a right angle as one of its angles, let the 
origin be the vertex of the right angle and let the positive parts 
of the x- and (/-axes contain the sides of the right angle. 

THEOREM 11,19 The medians of a triangle are concurrent in a 
point (eentroid) which is two-thirds of the distance from each ver- 
tex to the midpoint of the opposite side. 

Proof: Let A ABC he given. Select an ^-coordinate system in the 

plane of this triangle with the origin at A, with AB as the x-axis, with 
the abscissa of B positive, and with the ordinate of C positive. (See 
Figure 11-34.) Let a, h, c be numbers such that B = [Ga, 0), C = 
{&b f 6c), Let D s E, F be the midpoints of BC, CA, X73, respectively. 



i 


* cm*;) 




E(3b>3e/ J \Dr3o+36,3e> 

/ Nt'" A 


/uo.0} 


i-\3c.U) 0(60,0) 







Figur# U-34 



498 Coordinates in a Plane Chapter 11 

Then AD, BE, CFare the medians of A ABC. We must prove that AD, 
BE, CF are concurrent at some point P and that AF = JAD, BP m 
|BE, and CP = |CF. 

The midpoint of AB is F(3a, 0). We can express CF parametrically 
as follows: 

QP= {fcy) : x = 66 + {3a - 66)*, y = Gc + (0- 6c)*, < * < 1}. 

The point P on CF such that CP = |CF can be obtained by setting 
k = | in the parametric equations for CF. Thus 

x = bh + (3a - 66} • J 
- ($6 + 2a - 46 
= 2b + 2a 



and 



Therefore 



y = 6c + (0 - 6er) ■ § 
= 6c-4c 
= 2c. 



P= (2a + 26, 2c). 

Similarly, the midpoint of J5C is 

D = (3a + 36, 3e) 

and 

AD = {{s, • i = 4. (3c + 36 - 0)fc, 

i/ = + (3c- 0)*,0< fc < 1}. 

The point F on A D such that AF = %AD is obtained by setting k = -J . 

Tims 

s= (3a + 3/?)* | 

= 2« + 26 

and 

f -(*)*} 

= 2c. 
Therefore 

F = (2a + 2b, 2c). 

The midpoint of CA h E = (36. 3c) and 

BE = (fey) : x s 6a 4- (3fr - fia)*. y = + (3c - 0)*,0 < k < 1}, 

The point F' on BE such that BF' = §BE is obtained by setting k = § . 



11.9 Proofs Ustnf Coordinates 499 



Thus 



and 



Therefore 



x = 6a + (36 - 6fi) ■ | 
= 6a + 2fo - 4c 

= 2a + lb 

= 2c. 



P" = (2a + 2fc, 2c). 

Wc have shown that 

P = r = P" = (2a + 26, 2c), 

that CP = fCF, AP = \AD t and that BP = %BE. Therefore the medi- 
ans of a triangle are concurrent at a point which is two- thirds of the 
distance from each vertex to the midpoint of the opposite side, and trie 
proof is complete, 

In the second sentence of this proof we could have taken a t h t c as 
real numbers such that B = (a, 0) and C = {b t c). The resulting ex- 
pressions for the coordinates of D, E T l\ and P would have involved 
many fractions. We avoided these fractions by taking a, 6, G so that 
B = (0a, 0) and C = (65, 6c). 

You may feel that Theorem 11,19 would be easier to prove without 
using coordinates. It is possible to construct such a proof using the the- 
orems, postulates, and definitions that we have established before this 
chapter. You might be interested in trying to do so. 

As Indicated in the statement of the theorem the point of intersec- 
tion of the medians of a triangle is its ceutroid. In informal geometry 
we think of it as the balance point; it is the point where a cardboard 
triangular region of uniform thickness balances. In calculus the idea 
of moments of mass (extending the idea of weight times distance in 
teeter-totter exercises) is introduced and extended to develop a theory 
of centroids for plane figures. The centroid of a triangle is an example 
of a centroid as the concept is developed formally in calculus. 

Our last example of this section is a theorem that you will find help- 
ful in working some of the exercises at the end of the section. 

THEOREM 11.20 Let quadrilateral ABCD ivith A = (0, 0), 
B = (a, 0), D = (h> c) lie given. ABCD is a parallelogram if and 
only if C = (a + b> c). 



MM 



Coordinates in a Plane 



Chapter 11 



Figure 11-35 shows one possible orientation of Lhe given quadri- 
lateral ABCD in an xt/-plaiic. However, our proof depends only on A 
being at the origin and 8 being on the it-axis as given in the theorem. 
We must prove two tilings. 

h JfC = [a + b, c), then ABCD is a parallelogram. 
2. If ABCD is a parallelogram, then C = (a 4- b, c). 



D(b t e) 



a**y) 



^m 



5:q.,u; 



Figure 11-35 

Proof of h If C = (a + b, c), the slope of CD is 

c — c _ 0. = o 
a + b — b a 

Also, the slope of AB = ^ = 0, Therefore CD \\ AB. Since KB and UD 



ai"c horizontal segments (Why?}, we have 

AB = \a\ 
and 

CD = \a + b - b\ = a\. 

Therefore AB — CD and ABCD is a parallelogram. 

Proof of 2: If ABCD is a parallelogram, we must prove that 
C = (a + b r c). Since ABCD is a parallelogram, KB || t25. The slope 
of A J? is 0j therefore the slope of CD is 0. Let C = (x, y); then the slope 
of CD Ls 



= 0. 



a; — 



ITierefore y-c = 0andy= c. We have AB = CD (Why?) and 

AB = |d| and CD = \x - b\. 



Therefore 



\a\ = k - R 



11.9 Proofs Using Coord f nates 501 

Now, if a > 0, then x > b % and if a < 0, then * < b. (If a > and 
x < fc, or o < and x > fc t then C and B would be on opposite sides 

of AD and ABCD would not be a parallelogram.) It follows that 
a = x — b or that ac = a ■+■ fc. Therefore 

C — (a + h, c) 

and the proof is complete. 

EXERCISES 11.9 

Unless stated otherwise, use coordinates to prove the theorems in this set of 
exercises* Many of these theorems have appeared as theorems or exercises 
earlier in the texL We include them here since they can be proved easily 
using coordinates. 

1. Prove: 

THEOREM 11.21 If the diagonals of a quadrilateral bisect each other, 
then the quadrilateral is a parallelogram, 

(Hint: Let A = (0, 0). B = (o, 0}, C = (x, y), and D = {h, c) be die vertices 
of the quadrilateral and suppose die diagonals of the quadrilateral bisect 
each other. Show that x = a + b, y = a, and then apply Theorem 11,20.) 

2. Prove: 

THEOREM 11.22 The diagonals of a parallelogram bisect each other. 

(This is the converse of Theorem 1 1.21. By Theorem 11 -20, the vertices of a 
parallelogram may be taken as A = (0, 0). B = (a t 0), C = (a + h, c), and 
D = (h, c)+ Show that the midpoint of AU Is the same point as the midpoint 

,,] HI I) 

X Prove: 

THEOREM 77.23 If the vertices of a parallelogram are A = (0, 0), 
B = (a, 0), C =; (a + b t c), D = {b, c), then the parallelogram is a 
rectangle if and only if b = 0. 

(You must prove ( 1) if ABCD is a rectangle, then h = and (2) if b = 0, then 
ABCD is a rectangle.) 
4. Prove: 

THEOREM 11.24 The diagonals of a rectangle are congruent, 

(Let ABCD be the given rectangle. V$e Theorem 11,23 and prove 
AC m BD.) 



502 Coordinates in a Plane Chapter 1 1 

5, Justify the steps in the proof of the following theorem. 

THEOREM 11,23 If the diagonals of a parallelogram arc congruent, 
then the parallelogram is a rectangle. 

Proof: Let A = (0, 0), B = (a, 0). C = {a + b, c), D = (b t c) be the 
vertices of the given parallelogram. What theorem justifies our writing 
C = (a + fc, c)? 
We have 



Therefore 



and 
Therefore 



AC = BD. 

(AC)* = (BD) 2 , 

(AC) 2 = (a + &)» + ^ 2 > 

(«Z))8 = {& - o^ + c 2 . 



( fl + b) 2 + c 8 = (J> - «)2 + c a . 

Simplifying, 

a 2 + 2<ii> + L* 4- c 2 = b 2 - 2afe -f- a 2 + c\ 
2ab= -2ab t 



and 



4«/> = 0, 



Staee 4a 7^ 0* we can divide Ixith sides of the last equation by la, ob- 
taining h = 0. Therefore ABCD is a rectangle. 

6. Justify the steps in the proof of the following theorem . 

THEOREM 11.26 A rectangle is a square if and only if its diagonals 

are perpendicular. 

(See Figure 11-36.) By Theorem 11.23 
we may write A = {0, 0), B = (a, 0), 
C = (a, c), D = (0, c) for the vertices of 
the given rectangle. There are two things 
to prove. 

1. If AT X SB, then ABCD is a square. 

2. If ABCD is a square, tlicm A~C _ WD, 

Proof of 1: We arc given AT 1 UD. The slope of AC h r mid the 

slope of #D is -£—, Therefore 
— a 







— a 



11.9 Proofs Using Coordinates 503 

Therefore 

c 2 

- = _l ( fiS — fl a t an d f c f _ u , t 



-a* 

But 

|c| = BC and |« = Afl. 

Therefore BC = AB and A£GD is a square. 

Proof of 2: We ate given ABCD is a square. Therefore 

Afl = BC; \a\ = \c\ f and a- = c 3 , 
I lit r»j be the slope of AC and m-> be the slope of BD. Then 

Wi = — , 'Us = — -. and mirn^ = — -Q-, 
But 

— = l s .so mim? a — 1, 

Therefore W -I BB. 
7. Justify Ihc steps in the proof of the following theorem. 

THFOBKV 11.27 If the vertices of a parallelogram are A = {0, 0), 
B = (a, 0), C = (a + &, c), and D = {&, c), then the parallelogram is 
a rhombus if and only if a s = &+ c*. 

There are two things to prove, 

1. If a 1 = b s + c* then ABCD is a rhombus. 

2. If ABCD is a rhombus, then flS = 6 2 - A 

Proof of 1: We are given that a 2 = A* + e* so 

a V&* + C*i 
By tlie Distance Formula, 

ADm y/& + A 

Therefore AD = |of. Also, AB = a\, Therefore AB = AD and ABCD 

is a rhombus. 

Proof of 2; ABCD is a rhombus, so 

AD = AB 
and 

Therefore 

and the proof is complete- 



504 Coordinates in a Plans Chapter 11 

8, Prove: 

THEOREM 11.2H If the diagonals of a parallelogram arc perpen- 
dicular, then the parallelogram is a rhombus. 

(Hint: Let A = (0, 0), » = {a, 0). C = (a + b % c), and D = (6, c) be 
th e ve rtices of the given parallelogram. l,et mi and m% be the slopes 
of AC and BD, respectively. Using Theorem 11.27 of Exercise 7 and 
mi • m-t = — I , show that a 2 = b 2 -+- cK) 

9. Recall that a trapezoid is a quadrilateral with at least one pair of parallel 
sides which are called the bases of the trapezoid. The other two sides 
are called the legs of the trapezoid. The segment joining the midpoints 
of the legs is called the median of the trapezoid. Prove the following 
theorem. 

THEOREM 11.29 The median of a trapezoid is parallel to each of the 
bases and its length is one-half the sum of the lengths of the two bases. 

[Hint: Let A = (0, 0). B = {2a, 0), C = (2a\_2c), and_D a (2b f 2c) be 
the vertices of the given trapezoid. Then AB and CD arc the parallel 
bases. I*et E and F be the midpoints of AD and BC r respectively. Show 
that EF I AB t ET || CD, and that 

EF = fy\B + CD).) 

10. A trapezoid is isosceles if its legs arc congruent (See Exercise 9.) Prove 
the following theorem, 

THEOREM 11.30 A trapezoid is isosceles if its diagonals are 
congruent 

1L Prove: 

THEOREM 11.31 If a line bisects one side of a triangle and is parallel 

to a second side, then the line bisects the third side of the triangle. 

12. Prove: 

THEOREM 11.32 The midpoint of the hypotenuse of a right triangle 
is equidistant from the vertices of the triangle. 

(Hint Let A = (0» 0), B m (2a, 0), and C = (0, 2fc), where a, b are 
positive numbers, be the vertices of the given rigjht triangle.) 

13. Complete the proof of the following theorem. 

THEOREM 11.33 The lines which contain the altitudes of a triangle 
are concurrent. (Their common point is called the ortbocculer of the 
triangle.) (Sec Figure 11-37), 

Froof: Let A = (a, 0), B = (b, 0), C = (0, c)> and suppose that a < b t 
< c as shown in Figure 1 1-37. I jCI A', B\ C be the feel of the per- 



11,9 Proofs Using Coordinates 505 



i 


kj 






sm 


t 
/ 




-\ \ 


* \.mo) i. 


* * 


t 
t 
/ 


Ugam IU1T 



pcndfculars from A, B, C to BC, CA, AB, respectively. You arc to prove 

that AA', BB', CC' arc concurrent at some point F{xi, y{), (Note that 
(HO, 0) is the origin.) 



The slope of BC is 
b 



c-0 
0-h 



•v — r ■* r 

— — y. Since .'LA' 1 EC, the slope of 



A A' is — (Why?), Using the point slope form of an equation, we may 

express A A' as +_+ * 

AA' = {(x,t 7 }:i/=:^-a;.}. 

* — > 4 — > 

since (a, 0) is a point on AA'. We may express CC as 

CC={(x 1 tj) :* = <)}. 
Now write an equation for BB\ Let P(*i, 1/1) be the point of inter- 
section of lines AA', CO, and let F[x2, </?) b* f he point of intersection 
of lines BB', CC, Solve the equations for AA', CC "simultaneously" 
to find P[xi, t/i) an<l solve the equations for BB', CC simultaneously to 

find P(x 2 , ?/2)- Show that P = F and hence the lines AA\ BB'. CC are 
concurrent at P. 

14. Prove Theorem 11.18 by use of definitions, postulates,, and theorems 
studied before this chapter: that is, without the use of coordinates. 

(Perhaps you did this in an exercise of Chapter 10.) 

15, Use Theorem 11,38 and theorems, definitions, and postulates studied 
before this chapter to prove Theorem 11.29 without the use of 

coordinates. 

16* challenge Pitoni.KM. Prove Theorem 11.19 without die use of 
coordinates. 



506 Coordinates in a Plane Chapter 11 

CHAPTER SUMMARY 

In this chapter we used the idea of a coordinate system on a line to de- 
fine an .Tj/-coordinate system in a plane. We showed that there is a one-to- 
one correspondence between the set of all points in a plane and the set of all 
ordered pairs of real iwmlwrs. The key theorems in this chapter are: 

THEOREM 11.4 If F| = (x u yi) and P? = (** y 2 ) are any two 

points in an .vj/-plane, then 

PiPs = V(*i - «a) B + (m - W< 

THEOREM 11.5 If P = (x h ijt) and Q = (x 2 , ij 2 ) are any two distinct 
points in an jt]/-planc, then the midpoint X! of PQ is the point 



\ 2 ' 2 r 



TlIEOR EM 11.7 UA(xutfi) and Bix-t, ys) are any two distinct points, 
then 

AB = {x, y) : x = x t + k(xt - *i), y = y, + %o - y^fcisreal}. 

If jt is a real number and if F = (x, y) where ar = Xi + fe(*2 — Xi), 

u = yj - %a - yi)» &<»» 

AP = *(ArJ) and P^AJ? if it > 0; 

AP = _*(AJ9) and P£oppAR if Jt < 0- 

The formula in Theorem 1 1.4 is called the DISTANCE FORMULA. 
The formula in Theorem 11*5 is called the MIDPOINT FORMULA. 
The equations x z= x% + k(x$ — Xi) and y = tji ■+- fe(j/ 2 — #i) in The- 
orem 11.7 are called FARAMETRIC EQUATIONS for the line A~S, and 
k is called the PARAMETER. We proved that the converse of Theorem 
11.7 also holds; that is, if tt, h, a, d are real numbers, if b and d are not both 
zero, and if 

S = {{x* y) '. x = a + bk, y = c + <&-, fc is real), 

then S is a line. 

We defined the SLOPE of a non vertical line to be the slope of any one 
of its segments. We defined the slope of a non vertical segment with end- 
points Pi(*i, t/i', 'Wt**! Vs) *° l** 5 ■ We showed that two non vertical 

lines are parallel if and only if they have equal slopes and that two oblique 
lines are perpendicuilar if and only if the product of their slopes is — 1. Slope 
is not defined for a vertical line, but the slope of a hori7.ontal line is zero. 

We proved that if P|(*j, y{f and Psfra, y?) are any two distinct points 
on a nonvertical line, then 

* - *i _ V - tfi 
*a - *i J/2 - y\ 



Rsvlew Exercises 507 

is an equation of the line. We called this form of equation the TWO-POINT 
FORM for an equation of a line. We proved that a line with slope m and 
passing through H x i> tf i) has an equation of the form 

y - (/t = m(x - *i) 

and called this form the POINT SLOPE FOHM for an equation of a line. 
We proved that a line with slope m and {/-intercept h has an equation of the 
form 

y s= mx + b 

and called this form the SLOPE (/-INTERCEPT FORM for an equation of 
a line. We showed that an equation of a vortical line through ¥(%{, r/i) is 
X — K\ and that an equation of a horizontal line through P{x\, iji) is y = y\. 
Wc called the form Ax 4- By + C = the CpENERAL FORM for an equa- 
tion of a line. 

Finally, we showed how coordinates could be used to construct proofs 
for some geometric theorems and observed t hat, in some cases,, proofs using 
cord (nates are easier than those using previously established definitions, 
postulates, and theorems, 



REVIEW EXERCISES 

Graph each of the sets indicated in Exercises 1-10. 

1. {{*, y):x = 5, < y < 5} 

2. [(*f):r- -3, -2<C*<7} 

3. {(x, y) : x = 1 + 2Jfe, y = 2 + 3&, k is real} 

4. ((* y) : x = 1 + 2*, y = 2 + 3*. k £ 0} 

5. {(x, y) : x = -2 + fc, i/ = 1 - 2k k < 0) 

6. {(a*, t/) : x = *, y = 3 - 2k, 2 < k < «} 

7. {{*,«,) : l<x<4or2<y<5} 

8. {(x, y) i 1 < s < '1 and 2 < y < 5} 

9. {(x,y):y>2} 

10. {{x, y) : y = |x + 5} 

In Exercises 11-13, the endpoints A and R of a sequent AB are given. Find 
(a) the slope of AM (in lowest terms), (b) the midpoint of SB, and (c) the dis- 
tance AB. 

11. A = (2, 5} and ii = (-2,3) 

12. A - ;-L -3) and B = (2, -9) 

13. A = (2, -5) and B = (7, 7) 

14. A = (-2, 1) and B = (2, 3) 

15. A = (-3, l)andB= (7,1) 



508 Coordinates in a Plane Chapter 11 

16. Which segments in Exercises 11-15 are 

(a) parallel? 

(b) perpendicular? 

(c) congruent? 

17. Write, in general form, an equation of the line which contains the points 
P=(-h ^3) and = (2, -9). 

IS* Write, in general form, an equation of the line which has a slope of ^ 
and contains the point H = (—2, 1). 

19. Prove that the line of Exercise 17 is perpendicular to the line of Exercise 
18. 

20. Write., in slope (/-intercept form, an equation of the line with slope J 
arid ty-intercept 6, 

■ In Exercises 21-25, an equation of a line is given. In each exercise, (a) put 
the equation in slope {/-intercept form, (b) write the slope oi' the line, and 
:::) write the x- and {/-intercepts of the line. 

21. 3x + 2y m 12 

22. 2x - y = 7 

23. lox ~2lij = 7 

24. x + y = 

25. x — y = 

86, Stow that A SKM is a right isosceles triangle if S = {3, 4), X = { - 1, 3), 
andM = (-2, 1). 

27. Given A = (1, 0), B = (4, 3), express AS using set-builder notation and 
parametric equation*. 

28. Given A and B as in Kxerciso 27, express AB using set-builder notation 
and parametric equations. 

29. Given A and B as in Exercise 27, find the trisection points of AB r 

30. Write an equation of the vertical line through (2, 5). 

31. Write an equation of the horizontal line through % 5). 

32. Write an equation of the line through (3, —7) and parallel to the line 
with equation y = 3x -f 5, 

33. Write an equation of the line through (3, —7) and perpendicular to the 
line with equation tj = 3x + 5. 

34. Given the line with equation 5x - % = 00. write the equation of this 
line in slope {/-intercept form. 

35. Given the line of Exercise 34, write the equation of this line in intercept 
form. 

36. Given p = {(x, y) : x = 3} and q = {(x, y) : 2x = 6}, explain why 



Revww Exercises 



505 



Exercises 37-47 refer to the rectangle PQRS whose vertices are 
P = (-1, -4), Q = {5, -4), R = (5, .3), S = (-1, 3;. 

37, Fmd die midpoint oi PR. 

38, Find the midpoint of SQ. 

39, Show that PQ = $R r 

40, Show that PR = SQ. 

41, Write parametric equations for PQ, 

42, Write parametric equations for PR. 

43, Find A on Pit such that PA = 4PR. 

44, Find B on PR such that PB = fPR. 

45, Find C on opp M such that PC = J/»«. 

46, Write parametric equations for the line through 5 and perpendicular 
to PR. 

47, Write parametric equations for the line through R and parallel to SQ. 



48, 



Let die trapezoid ABCD have vortices A = (0, 0), B = (2a, 0), 
C = (2(1, 2c\ and D = (2/?, 2c), as shown in the figure. Let a, b, c, d be 
positive numbers such that b < d < a. Let E and F be the midpoints of 
AD and BC, respectively. Let EF intersect AC at P and SD at {). Show 
that P is the midpoint of AEL that Q is the midpoint of ED, and that 
P£ = {{AB - CD). 



P(M),2e) C(2d,2e) 




£(2b,0) 



49. Prove, using coordinates, that the set of all points in a plane equidistant 
from two given points in the plane is the perpendicular bisector of the 
segment joining the given points. (Hint: Let the two given points in an 
xy-plane be A = (— a r 0) and B = (a, 0). Then the {/-axis is the per- 
pendicular bisector of AB in the given .try-plane.) There arc two things 
to prove. 

(a) If P(x, y) is on the y-axis, then AP = PH. 

(b) If AP = PR and P is in the .ty-plane, then F is on the y-axis. 

50, challenge piiohlum. Prove that the urea S of a triangle whose ver- 
tices are A = (x\ t tj\), B = (X2, y2), and C =a [xz, y$) is given by the 
formula 

s = i^i^a + *»03 4- * 3 yi ~ *lft - x *y* - *fc§fll* 




Bradley Smith/Photo Researcher!, 



Coordinates 
in Space 



12.1 A COORDINATE SYSTEM IN SPACE 

In Chapter 3 we introduced the fundamental idea of a coordinate 
system on a line, or a line coordinate system, or a one-dimensional co- 
ordinate system, as it is sometimes called. Tn Chapter H, we defined a 
coordinate system in a plane and called it an *j/-coordinate system. An 
any-coordinate system is a one-to-one correspondence between all the 
points in a plane and all the ordered pairs of real numbers. Each point 
has two coordinates. An ^-coordinate system Is a two-dimensional co- 
ordinate system. In this chapter we introduce the idea of a coordinate 
system in space and called it an xi^-coordinate system. In this system 
each point of space is matched with an ordered triple of numbers. It is 
•a three-dimensional coordinate system. 

Let a unit segment and die distance function based on it be given. 
All distances will be relative to this unit segment unless Otherwise 
indicated. 

Let OX and OY be any two perpendicular lines and let OZ be the 

unique line that is perpendicular to each at their point of Intersection, 

Let J, J, K be points on OX, OY, OZ % respectively, such that 

01 = OJ = OK = 1. 



512 Coordinates in Space Chapter 12 

On OX there is a unique line coordinate system with O as Origin 
and / as unit point. On Or there is a unique line coordinate system 
with O as origin and / as unit point. On O'Z there is a unique line co- 
ordinate system with O as origin and K as unit point. We call these 

coordinate systems the x-eoordinate system on OX, the {/-coordinate 

system on OY t and the ^-coordinate system on Sz. We refer to OA\ 

OY, OZ as the x-axis, the (/-axis, and the *-axis, respectively We refer 
to them collectively as the coordinate axes. 

The plane containing the x- and y~axe$ is called the xy plane, The 
plane containing the x- and s-axes is called the j/z-plane. The plane con- 
taining the y- and s-axes is called the r/z-plane. We refer to these three 
planes collectively as the coordinate planes. 

From die theorems regarding parallelism and perpendicularity in 
Chapter 8 it should be clear that all lines parallel to the .s-axis arc per- 
pendicular to the K^-plane, that all lines parallel to the (/-axis are per- 
pendicular to the xz-plane, and that all lines parallel to the X-axis are 
perpendicular to the 1/2-plane, It should also be clear tit at all planes 
parallel to the xy-plane are perpendicular to the z-axis, that all planes 
parallel to the xz-plane are perpendicular to the [/-axis, and that all 
planes parallel to the i/s-plane are perpendicular to the x-axis. 

Figure 12-1 suggests the x-, i/-, and ^rcoordinate systems, Hie 
parts of tlie axes with negative coordinates are shown by dashed lines. 
They are not "hidden" from view by the coordinate axes in the figure. 
However, they are hidden from view by die coordinate planes, "llius 
the negative part of the x-axis is behind the i/s-pkne; the negative part 
of the c/-axis is hidden by the ace-plane; the negative part of the c-axis 
is hidden by tlie X(/-planc, In drawing pictures you should use your own 
judgment about whether a dashed segment is better than a solid one. 



Figure IS- 1 




12.1 A Coordinate System in Space 513 



In Figure 12-1, the positive parts of the x-, y- 3 and £-axes are ar- 
ranged as the thumb, forefinger, and middle finger, respectively, of a 
right hand when it is held as suggested in Figure 12-2* An .ti/5-coordi- 
natc system with axes oriented in this manner is called a right-handed 
coordinate system. If the unit points on the axes are selected so that 
the positive parts of the x-, y-, z-ax&s con form to the orientation of the 
thumb, forefinger, and middle finger of the left hand when the thumb 
and forefinger are extended and the middle finger is folded, the xyz- 
coordinate system is called a left-handed coordinate system. The fig- 
ures in this book arc for a right-handed system. 




Figure 13-2 

There is an xy-coordinate system in the xy-plane determined by 
O, /, / as in Chapter 1 1 . This system is a one-to-one correspondence 
between the set of all points in the xy-plane and the set of all ordered 
pairs of real numbers. Similarly, there is an xs-coordmate system in the 
xs-plane and a ys-coordinate system in the y^-plane. 

T^et F be any point in space. 
Figure 12-3 shows P as not 
lying on any of the coordinate 
pianos. However, the following 
discussion which leads to the 
definition of an xf/z-eoordinate 
system applies to any point. 
For special positions of P some 
of the labeled points thai are 
distinct in Figiire 12-3 may 
not be distinct. For example, 
P and F IU may be the same 




point. 



Figure 12-3 



514 Coordinates in Space 



Chapter 12 



Let P w P«, V vt be the projections of P on the xt/-plane, the xs-plane, 
and the i/s-plane, respectively. I-et a XUt a gs , a^ be the planes through 
P and parallel to the xy-plane, the xs-plaiic, and the i/z-plane, respee- 
lively. Then a rv contains P t P rs , and P us ; fife* contains P, F m and F m % 
and %, contains P, P Mt and /', 



••..,. 



Let @ n py, () t be the points in which a lw , a X!t a^ intersect the 
x-axis, the y-axis, and the s-axis, respectively. We are now ready to set 
up an .rr/s-eoordinate system. 

The ^coordinate of P is the ^.--coordinate of Qy, the y-coordinate of 
P is the ^-coordinate of Q^ the s-coordinate of P is the ^-coordinate of 
Q z . Wc write 

P = (a, 6, c) or P(a, Z? } c) 

to indicate that the *-, y-> s-coordi nates of P are a, b, c, respectively. 

If (a, h } c) is any ordered triple of real numbers, then there is one 
and only one point P such that 

P=P( a ,b>c), 

It is the intersection of three planes, one parallel to tbe xs-plane and 
cutting the *-axis at the point whose x-coordinate is a, etc. 'The cor- 
respondence between the set of all ordered triples of real numbers 
and the set of all points is a one-to-one correspondence. For if {«, b, c) 
and (rf, e t f) are different triple*; of numbers, then one or more of the 
following inequalities must hold: 

a^d, 6#e, c=£f. 

Suppose, for example, that a =£ tl Then P(«, b t c) and R(d> e, f) He in 
distinct planes parallel to the i/z-plane and therefore P^R. 



Definition 12.1 Given an t-axis, a (/-axis, and a zf-axis, the 
one-to-one correspondence between all the points in space 
and all die ordered triples of real numbers in which each 
point P corresponds to the ordered triple (a, h, c) where a, b, c 
are die Xh, y-, ^coordinates, respectively* of P is die xyz- 
coordinate system, 



Since the correspondence between points and triples is one-to-one, 
a system of names for the triples is a suitable system of names for the 
points. Thus (5, 6 r -3) is an ordered triple of real numbers. It is also a 
point. It is the point whose x-, y- f and s-coordinates are 5, 6, and —3, 
respectively. 



12.1 A Coordinate System in Space 'j l :■ 

Example 1 If 

A = {(*, y, z) : x = 2. tj - 3, * = 4}, 

then 

A m {(2, 3, 4)}, 

that is, A is the set whose only element is the point (2, 3, 4). 

Example 2 If 

B={{x t t,,z):x = -2.y = 3}, 

then J3 is the set of all points (a; y, 5) such that the x-coordinate is —2 

and the y-coordinate is 3, that is, £ is the line through ( — 2, 3, 0) and 

parallel to the .-7 -axis. 

Example 3 If 

C = {{*, y, z):y = 7}, 

then C is the set of all points whose ^-coordinate is 7, that is, C is the 
set of all points that are 7 units to the right of the acs-plancj hence C is 
the plane that is parallel to the xz-planc and 7 units to the right of it. 

Example I 

D = {(*, y, z) : x = 2 and y = 3}. 

The use of "and" !n this set-builder notation is the same as the use of a 
comma between statements of conditions in a set-builder notation. In 
other words, 

D = i(x i y i z}:x = 2,y = 3} 

is the set of all points (2* 3, z\ that is, the line parallel to the s-axis and 
passing through (2, 3, 0). 

Example o 

E = {(x, y, z) : x = 2 or y s 3}. 

Clearly* E is not the set D of Example 4. If you do not understand 
dearly the distinction between the use of "and" and "or" in a set- 
builder notation, you shoxild review Chapter 1. E is the set of all points 
with ^-coordinate 2 or ^-coordinate 3, hence the set of all points lying 
either in the plane x m 2 or in the plane y = 3. Tn other words, £ is the 
union of two planes, each parallel to the s-axis, one of them parallel to 
the yx-planc and 2 units in front of it, the other parallel to the xz-plane 
and 3 units to the right of it, assuming that the axes are as in Figure 
12-1, 



516 Coordinates in Spac* Chapter 12 

Example 6 

F = {(x, y,z):x=l + Zk t y = 2^ 3k t k is real}. 
Now 

{(*, y) : x = 1 + 2k, y = 2 + 3k, h is real} 

is a line I in the xt/-planc with slope | thai pisses through (1> 2). The 
set-builder notation for F places no restriction on the ^-coordinate. 
Therefore, if {x, y) is any point on I and z is any real number, then 
(x, y, z) is a point of ft Conversely, if (*, y, z) is any point on F, then 
(x, y) is a point of /. Think of /' as the set 

{(a; j/,2);a;=l + 2*,j/ = 2 + 3k, z = 0, k is real}. 

Then V is the same set as I If we think of a point of I, we think of it as a 
point (x, y) in an xy-plane. If wc think of the same point as a point of /'» 
we think of it as a point (x t y, 0) in an xip-spacc. Although the names 
of the points are different, the sets I and I' are the same. The graph of 
the set V is shown in Figure 12-4, 




Figure 12-4 

Now V is a part of ft Indeed, F is the set I' and all points directly 
above it and all points directly below it. It may be helpful to think of it 
as the union of all linos parallel to the *-axis that pass through a point 
of V. But any way you look al F t you really have not seen it unless you 
recognize it as a plane, the piano through /' and parallel to the *-axis. 

Example 7 

G = {(x t y,z):z<3}> 

G is the set of all points (*, y, z) whose z-coordinate is less than 3. Now 
z = 3 is an equation of the horizontal plane, say a, that is 3 units above 
the xc/-pkne. Then C is the halfspace that is the underside of a. 



12.1 A Coordinate System In Spact 517 
8 

H=[(x t y 1 z):l< i z< 3}. 

H is a slab or zone of space bounded by two horizontal planes. H is the 
union of the horizontal planes with equations 5=1 and z = 3 and 
all of the space that lies between them. 

Example 9 

I 8 {(*, tj,z):x* + y* = 25 t z = 0). 

1 is a cirde in the xy-plane. Its center is the point (0, 0, 0). Its radius is 5. 

Example 10 

/ is a solid right circular cylinder. Its axis is a part of the s-axis. Its 
radius is 5. Its lower base lies in the xi/-plane. Its height is 6. 

Example 11 

K = {(*, y, z) : x = 2 and x = 3}. 

There is no number x such that x = 2 and x = 3. Therefore there is 
no point (x, \j, z) such that x = 2 and x = 3. Therefore K = 0, the 
null set. 

Example 12 

L = {(*, y, z) : x* - 3* + 2 = 0}. 

Now x> - 3* + 2 = is true if and only if (x - l)(x - 2) = 0; hence 
if and only if x = 1 or x = 2. Therefore L is the union of two planes, 
each parallel to the t/z-plane, one of them 1 unit in front of it, Ihc other 

2 units in front of it. 



EXERCISES 12.1 

In Exercises 1-35, a set S of points is given in set-builder notation In each 
case, descrilw the set S in words, assuming that the axes appear as in 
Figure 12- 1. {Hint: For Exercises 20, 21. and 22, compare with Example 12.) 

L $={{x t y.z):x = 0) 
%S={(x,y.z):y = Q} 

3. S = {(x, y. z) i z = 0} 

4. S= ((x,y,2):x = 0,y = 0) 



518 Coordinates in Space Chapter 12 

5. S = {<x s y, 4 19 m %* = 0) 

6. S= {(x,y, £ )ry = 0»z = O) 

7. S = {(x, y, s) : x = Q, y = Q, s = 0} 

8. S={{x,y,- S >;x=l,y = 2,z=2} 

9. S = {(*, y, z) : x = ly = 2, ] < * < 5} 
10. S={(*,y,z) : a: = I, 1 <y^3,* = -I) 
It S = {(x, y, 2) : * £ 1, y = 3, z = 1} 

12. S = {(*, y, s) : y < 3} 

13. S = {{x, y, z) : x = 3, tj < 3} 

14 t S = {(x, y, 2) : 1 <, x < 2, 3 < y <, 5, -1 < z < 3} 

15. S = {{*, £/, z) ; x £ 0> y > 0, * > 0} 

16. S = {{x, y, z):x> 370} 

17. S = {(x, y,z):x = ly = 2) 

18. S ss {(x, y, jj) ; x = 1 or J/ = 2} 

19. S= {{x,y r z) : x = 5orx = 7} 

20. S = ((x, y, 2) : s* + 3s + 2 = Oj- 

21. S = {(x, y, z) i & + 2* + 1 = 0} 

22. S = {(x, y, s) : x 2 = 16} 

23. S={(x,y,s}:x 2 + y 2 = 25) 

24. S = {(x, y, z) : x 2 + y* = 5, * - 5} 

25. S= {(x,y, s ) :i/ = * + 3,5 = 4} 

26. S = {(x, y, s) : tf s x} 

27. S = ((x, y, *) : z = y} 

2S. S = [<x, y, z) : 3x + Ay = 12, x £ 0, y > 0, 1 < 2 < 2} 

29. S={(^, / ^):^ll == J^L? tS = 0} 

30. S = {(x, y, z) ■ 2±1 = £z-L. y = } 

3L S = {(x r j/,*):.t = 1 + 2k, y = 2 - ft, z = 0, 1 < Jfc < 2} 

32. S = {(x, y, z) : x = 1 + 2*, y = 2 - Jfc, 1 < * < 2) 

33. S = {[x, y, g) : x= + y* ^ 2 5, s = 3) 

34. S = (fx, y, z) : y 2 + z* < £5. x = 3) 

35. S = ({x,y,s):x#0.y.^O\s ? feO} 

■ In Exercises 36-50, use set -I milder notation to express the set 5. 

36. 5 is the xy-planc, 

37. S is the xs-plane. 

35. S is the j/*-p1ane. 

39. S is the x-axis. 

40. S is the y-axis. 



12.2 A Distance Formula 519 



41. S 

42. S 

43. S 

44. S 

45. S 

46. S 

47. S 

48. S 

49. S 

50. S 



is the plane through (2, 6 f 7) and parallel to the yz-plane. 
is the plane through (2, 6, 7) and parallel to the xs-plane. 
is the plane through (2, 6, 7) and parallel to the xy-pfane. 
is the line through (2, 6, 7) and perpendicular to the t/2-plane. 
is the line through (2, 6, 7) and perpendicular to the xs-plane. 
is dio line through (2, 6, 7) and perpendicular to the xy-plane. 
is the line in the xr/-plane which contains (2. 3, 0) and (3, 7, 0), 
is the ray AB, with A = (5, 3, 0) and B = (4, 6, 0). 
is the segment Fp, with P s (2, 1, 0) and Q = (0, -7. 0). 
is the segment RS, with R = (2, I, 1) and S = £0, —1,1}, 



12.2 A DISTANCE FORMULA 

In this section we develop a formula for the distance between two 
points expressed in terms of their coordinates. We begin by consider- 
ing an example. 

Example 1 Let A - (3, 2 t 1) and B = (5, 4, 2). (See Figure 12-5.) I .el 
A a = (3, 2, 0), B t = (5, 4, 1), B 2 = (5, 4, 0), B 3 = (0, 4, 1), and 
B* = {0. 4, 2). Then AA2B2B1 is a rectangle and ABi = Ajfiz. Also 
BBiBzB* is a rectangle and B t B = BsB 4 , 




Bjl Figure 1S-5 

To find A2B2 we use the Distance Formula and the x- and ^-coordinates 
of A 2 and 13* In the xy-eoordinalc system. As = (3, 2) and B 2 = (5, 4). 
Therefore 

A 2 B 2 = V(5 - 3)* + (4 - 2) a = \/8 = 2V2. 



520 Coordinates in Space Chapter 12 

To find B3B4 we use the Distance Formula and the y- and subordinates 
of B3 and B+ In the ^-coordinate Systran B 3 = (4, 1) and B 4 = (4, 2). 
Therefore 

B S B^= V(4 - 4)z + (1 - 2)2 = VT= L 
Now BBi is perpendicular to the xy-pkne. It is also perpendicular to 
every plane parallel to the ay-plane. In particular, S?i is perpendicular 

to the plane with equation !■! Then mh & perpendicular to every 

* — * ■ — * 
line in that plane through tfj. Therefore BB t _L AB t and AABiB is a 

right triangle. It follows from the Pythagorean Theorem that 

{AB)* a (AB l )2 + (B^)* 
(AS)* = (A 2 i3 2 ) 2 + (£ 3 B-i) 2 
(AB)* = (V§) 2 + I 2 
(AB)S = 8+1 = 9 
AJ3 s 3 

Following the procedure used in this example we shall prove the 
next theorem. 

THEOREM 12.1 {Distance Formula Theorem) The distance 
between F{x lt y it z{) and Q{x& y% z 2 ) is given by 

PQ = Vta - *i)» + (sf! - yir + (* - *)*. 

jftoo/* Let P(*i, y lp si) and ()(x a , 1/2, afe) be given. Let 

P x = (xi, yi f 0) 
Qi = (*fe gfe 21) 
Qz = (*a, y 2 , 0) 
<?a = (0, y 2 , xi) 

Figure 12-6 shows these seven points as distinct points. Depending on 
die values of the coordinates, the points may turn out not to be distinct. 
For example, if zj = 0, then P = P x . 

We proceed to find PQ in terms of the coordinates of P and Q. If F 
and Q\ are distinct points, then they lie on a plane parallel to the 

u/-plane and PQ ± is parallel to the xy-plane. Therefore 

(1) PiQtQiP Is a rectangle, 
or (2) P = Ft and ft = Q 2t 

or (3) F = Q L and P, = £ z . 



12.2 A Distance Formula 521 




Figure 12-0 



In all three cases it is true that 

(FQ 1 f = ftQtf = fa - *i)» + (y* - yi) 2 - 

Similarly, (I) QQ1Q3Q4 is a rectangle, or (2) Q = Q 4 and £i = Qz, 
or (3) Q = pi and p 4 ■ Qn In all three cases it is true that 

If 9 = p Xi then % = s 1? (sz — zi) 2 = 0, and 

Hf = (PQtf = {x 2 - xj* + (tfz - y,) 2 + (z 2 - ft) 2 . 



If P = p l5 then 12 = afe $2 = 5fo (*2 - *i) 2 = 0. (1/2 - gfo? = ft 
and 

(FQ)2 = (p!p) 2 = (ss - *i) 2 + (t/2 - f/0 2 + (*b - *i) 2 - 

If Q ^ p x and P 7*= pi, then ^5, 1 FQi> mid it follows from the 
Pythagorean Theorem tliat 

(Pp) 2 = (FQtf + (WrP, 

(Pp)2 = fe - X i)» + (y 2 - yi) 2 + fo - s,} 2 . 

We have now exhausted all cases. For if P and Q are any points and if 
pi is related to P and Q as indicated in the first sentence of this proof, 
thenP = Pi; or Q = Qn or P, Q, Q\ are three distinct points. We have 
proved in all of these cases that 

(Pp)2 - fa _ Xl ) 2 + (tfz - yi) 2 + (^ - *i) 2 - 
Therefore 

pq = vte - xi) 2 + (ys - yi? + te - »i)*"- 



522 Coordinates in Space Chnptsr 12 

EXERCISES 12,2 

■ In Exercises 1-5, find the distance between Panel Q. 

LPs (0, 0, 0), Q = (3, 4, 5} 

2. F=(2, -1.7), Q= (2, 1.-7) 

3. P=(7, 8,1), p = (0 t fl,0) 
4.P={0,0,5),@ = (3,4,0) 

5. P=(a -J,5),(? = (1, -1,6) 

6* Use the Distance Formula to show that die following points are 
collinear: A m (-1,5, 2), B = (5, 7, 10), C = (2, 6, 6), 

7. Use the Distance Formula to show tliat the following points are not 

collinear: A = (0, 0, {)), B = {5, 3, -4}, C = (15, 10, - 12). 
a If A = (L 3, -2), H = (2, 7, 1), C as (2, 6,0), is A ABC a right triangle? 
9. IfA = (5,3,2), B a (7,8, 1), C = (5,2, -3), is AABCa righttriau I ' 

10. Find the perimeter of &DEP if D = (0, % 0), E = (0. 6, 8). 
F = (10, 6, 8). 

■ In Exercises 11-14, find tl*e distance between F{2. — 1, 7) and the given 
plane a. 

11. a = ((x, y. s) : x = -5} 13. a = {(x, y, =) : z = 0} 
IS. a={{x,y,z):y = 0) 14. a : = {(x, y, z) : y = ~1) 

■ In Exercises 15-18, find the distance between the line / given by 

I = {(x, y r z):x = 3, y = 4} 

mid the given plane a which is parallel to it. (The distance between u line 
and a plane parallel to it is not del ined in the book. 'Write a suitable definition 
for tliis distance.) 

15, a = {(r } y t z) : x = 0} 17. a = [(«, y, z) i ; ij a 6} 

16. a — {(x, y s z) : i = —5) 18. a m {(x, y, s) : tj = 4} 

■ In Exercises 19-23, sketch a figure to represent the given set. (The graphs 
will be three-dimensional.} 

19. A = {(*, y, s) : 1 < * < fi, £ < y < 4, < ;= < 10} 

20. B = {(x, y, 5) ; 1 < a: < 2, 6 < y < 8. 4 < 2 < 6} 

21. C = {(x,tj,z):l<x< 2, 10 < y < 12, < = < 10} 

22. D = {{x, y> z) : 1 < x < 2, 14 <, y < 16, 4 < z < 5 or 6 < a < 7) 
S3. £ = {(*, y, sj : 1 < x £ 2 t 18 < «/ < 20, < z < 10} 

24, challenge problem. On one large sheet graph the union of the sets 
in Exercises 19-23. 



12.3 Parametric Equations for a Una in Space 523 



12.3 PARAMETRIC EQUATIONS FOR A LINE IN SPACE 

Before studying Section 12,3 you should review the work on para- 
metric equations in Chapter 11 . By use of a proof similar to the one for 
Theorem 1 1.7 the following theorem can be established, A proof of this 
theorem is not included here. 

THEOREM 12.2 If P(* 1t yu -i) and Qix 2 , ua, z 2 ) are any two dis- 
tinct points, then 

x = xi + k(x z - x\) t 
W$ = (x, y, z) : y = tfi + k(tf2 - yi)> and k is real 

z = si ■+■ kizz — «i), 

If R a {a, b, c) t where 

a = x-i + k{xz — x\), 
h= yi + %j - tji), 
c = zi + k(2t - *i), 

is a point of FQ t then 

It is the point P 

R£PQ and PR = k * Pp 

H € <W P£ and PR = -Jfc ■ P@ 

Example 1 Given P = (0, 0, 0) and Q = (3, 1, -2), find the point R 
on P$ such that PR = 3 ■ FQ and the point S on opp PQ such that 

Solution; 

P$ = {(x, U, z) : x = 3fc, y = *, * = -2fc, fc is real} 

To get fl % set fc = 3. Then R = (9, 3, -6). 
To get 5, set k = —3. Then S = (-9, -3, 6). 

Bampfe 2 Given A = (5, 1, -2) and £ s (7, 3, -2), find the mid- 
point M of AB. 

Solution; 

£jf = {(*, y, s) : x = 5 + 2fc, y = 1 + 2*, 2 = -2,kis real}. 
The ^-coordinate of A is 0; the fc-eoordiuate of B is 1 ; of M is £. 
M = (5 + 2 ..^ 1 + 2 - J. -2) = (6, 2, -2). 



if 


k = 0, 


tf 


Jt>0, 


if 


k<0. 



524 Coordinates in Spact Chapter 1 2 

Example 3 Given £ = (5. 7, 9} and F s (— 2, 3, —4), write para- 
metric equations for Et. 

Solution: x - 5 - 7Jt 

y = 7 -4k 
s = 9 - m 
*£ 

Example 4 Given 11 = (5, -3, 14) and J = (-2, 5, - 1), write para- 
metric equations for lit 

(1) with a parameter Jfc so that A: = at H and & = 1. at 1. 

(2) with a parameter h so that /« = at I and fa = 1 at H. 

(3) with a parameter t so that f = at 7/ and t = 10 at L 

Solution: 

(1) x= 5 - 7* (2) * = -2 + 7ft (3) x = 5 - 0.7* 
y = -3 + Bk u = 5 - 8/i y = -3 + 0.8* 

x= 14-15* S=-1 + I6ft z= 14-1.51 

G < fc < 1 < « < 1 < f < 10 

Example 5 Given J = (5,1, 3) and K s (7, —1, 5). find an equation 
for the perpendicular bisecting plane of JK. A simplified equation is 

desired. 

Solution* Let a be the perpendicular bisecting plane of JR. Then 
P(x> y, s) is a point of a if and only if 

JP a EK, 

tff)» - (IX)*, 
(* - 5)2 + (y - 7)2 + (a - 3)2 = (x - 7) 2 + (y + If + (s - 5)2, 
x- 2 - lOx + 25 + y« - 14?/ + 49 + s* - 6s 4- 9 

= x 2 - I4x + 49 + y 2 + 2y + 1 + s 2 - 10s + 25, 
- lOx + 14x - Uy - % - 6s + 10s 

+ 25 + 49 + 9 - 49 - 1 - 25 = 0, 
it - 16y + 4z + 8 = 0, 
x — 4ij -r z + 2 = 0. 

Then x — Ay + s + 2 = is "an equation of a" This means that 

a = {(x, y, s) : x - 4y + s + 2 = 0}. 



12.3 Pwsroetrk Equations for a Line in Space 525 



EXERCISES 12.3 

In Exercises 1-4, find the coordinates of the point P satisfying the given 
condition. 

L A = (0 T 0, 0), B = (3, 3, 3). P€ 3$. AP = §>AB, 

2. A = (0,0,0>,B = (3,3,3),P€ W a5,AP = * 3 *AB 

3. A = (5, -2, 6). £ = (1, -3, 2), P C HA» AP = 2 ■ AD. 

4. A = (-3, 2, 5), fl = (5, 0. - 1), P € XB t AP = FB. 

In Exercises 5-10, A = (0, 0, 0), B s (6, -3, 12), C = (-9, 18, 9). 

5. Find D, the midpoint of J3fl 

6. Find B, the midpoint of ACT, 

7. Find F, the midpoint of A B. 

8. Find Gi, the point that is two-thirds of the way from A to D, 

9. Find G2, the point that is two-thirds of the way from B to E. 

10. Find G& the point that is two-thirds of the way from C to F. 

11. Given S = {(x, y , z) : 3x + 4i/ + 5s = 50), which of the following 
points are elements of S: A(5 t 5, 3), fl(G, 0, 5), C{- 1, -2, 12}? 

12. challenge problem. Find tlie point on the T-iutis that is equidistant 
from A{2, 3, 1) and B{7, 2, 1). 

13. challenge problem. If S is the set of all points each of which is at a 
distance of 7 omits from the origin. End an equation (in simplified form) 
for S. What is a set of points like S usually called? 

14. challenge pboblem. If T is the set of all points, each of which is at a 
distance of 3 units from the point (3, —1.2), find an equation (in simpli- 
fied form) for T. 

1.5. CHALLENGE PROBLEM. (Jiveil 

S = {(-Y, y, a) : & + f 4- z 2 = 125}, 

T = {(x, y, z):x = 5}, mid Q = (5, 0, 0), 

prove that all points P in the intersection of S and T are at the same dis- 
tance from Q. 
16. challenge problem. Let planes u and and points, O t A, A Ir B, B^ 
be given as follows: 

/? = {(*. y, z) : j 4- y + s = 9}, 

O = {0, 0, 0), A = (2. 4.2). Ai = (6.0, 0), 

B = (3, 3, 3). Bi = (9. 0, 0). 

(a) Show that A and A i are points; of a. 

(b) Show that B and Bi arc points of ft. 



526 Coordinates « n Space Chapter 12 

:V) Show that UA 1 ZKl. 

(d) Show that UB 1 SSI, 

(e) Find a point Aa in a such that A, A], A* are noneollinear. 
(i) Find a point H* in /? s uch llial B» B lt Ba are non colli near. 
(g) Show that UA ± AAg, 

(h) Show that US 1 SSi. 

(!) Show that DA 1 a. 

(j) Show that DZJ 1 j8. 

(k) SIiow that 0, .4, iB are colHnear. 

(1) Find the distance between a and /i. 
17. CHALLiiscjK i-itonr.nM. I.et planes a and /3 and line / lie given as 

follows: 

a = {(x t y, z) : x + y + 2,? = 8}, 
= {(*, y, s) : x + y + 2* = 2}. 



= [{x, i/, z) : x = y = ^H 



(a) Find the point A in which / intersects a. 

(b) Find the point B in which / intersects /?. 

(c) Prove that A~B 1 a. 

(d) Prove that AB _ fi, 

(e) Find tlie distance between a and /?. 



12.4 EQUATIONS OF PLANES 

In Chapter II we used coordinates to write equations of lines. For 
example, 

3* + Atj +■ 7 = 

is an equation of a line in an ty-plane* We say that 

3x + 4y + 7 = 
is a linear equation in y and t/. The equation 

ax 4- ^ 4- G = 

is called the general linear equation in x and y. It is natural to extend 
this terminology and to call 

ax + hy -h cz 4- & — 

the general linear equation in x, !/, z. In this section we show that, if an 
xyz- coordinate system has been set up, every plane has an equation of 
the form 



ax + by + '* + d — 



12.4 Equations of Plane* 527 

with «, h, & d real numbers and with a t b, c not all zero, and conversely, 
that every equation of this form is an equation of a plane. What is the 
situation ifa,h t and c are all zero? More specifically, what is the graph 
of Qat + % + G*+I = 0? Of Ox + Qy + CMr + = 0? 

We begin with several examples, assuming that an xy ^-coordinate 
system has l*ecn set up in space. 

Example t IaH = (0, r 0), P = (3, 4. -6), and let a be the plane 
that is perpendicular to UP at P. We shall exjyrefa a in terms of the co- 
ordinates of the points on it using set-builder notation, Now 

OP = (fr tf % z):x = 3k>y = 4*. z = -6*, k > 0}, 

I ei: O I h 1 1 1 1 point on OP that is twice as far from O as P is from O, To 
find the coordinates of Q, set k = 2, then (J = (6, 8, - 12), P is the 
midpoint of UQ t and a is the perpendicular bisecting plane of U%). 

The perpendicular bisecting plane of a segment is the set of all 
points equidistant from the endpoints of the segment. Using the Dis- 
tance Formula, we find that the distance between O(0, 0, 0) and (x, y, z) 
is 

V* 2 + yf + a? 

and that the distance between P{6, S, — 12) and (x t y t z) is 

V(* - 6)2 -f- {y - 8)2 + ( S + 12)3. 

Therefore 

« = {(X, y, Z) : # + 7 + ^ 

= v'«* - 8)* + {y - 8) 2 + (* + 12) 2 }. 

a = {(r, p, c) : s* + y 2 + s 2 

= x 2 - 12% + 36 + t/2 - 101/ + 64 + z 2 + Uz + 144) f 

« = {(*, y. s) : Hx + 1% - 24s - 244 = 0}, 
« = {(*. y» s) : 3* + 4y - 6* - 61 = 0). 

Example 2 Let O = (0, 0, 0% P = (3, 4, - 6), and let /? be the plane 
that is perpendicular to OF at O. Note that /? is parallel to the plane a 
of Example 1 and that a. docs not contain O, but that fi does. Now 



OP = ((*, r/ s z):x = 3k, y = 4Jt, z = -6k, k Is real}. 

Let R be the point on opp OP such that OH = OP To find the coordi- 
nates of R, set k = — 1, then K sb (—3, — 4, 6), and O is the midpoint 
of BP and $ Is the perpendicular bisecting plane of KP. 



528 Coordinates in Spac* Chapter 12 

/? is the set of all point 4 ! (x, y, z) each of which is at the same distance 
from (3, 4, -6) as it is from (-3, -4, 6), Therefore 

P = (fe .'/> A = V(« - 3? + (y - 4)» + fr + 6)2 

= ^ + 3)> + (y + 4)* + («-6)S} t 

fi = {(*, y, «) ■ : x s - &x + 9 + i/2 - 8y + 16 4 z 2 4 12s +- 36 

= x 2 + 6* 4- 9 + !/ 2 + 8y + 16 + s£ - 12s 4 36}, 
p a ((* //, 4 : -12* - 16y + 24* = 0}, 
= {(*, t h Z) ; 3* 4 4y - 62 = 0}. 

ficample 3 Let y he the following set of points in .fys-space: 
y = {(*, y, *) : * - 2y + 3a-4 = 0}. 

We shall show that 7 is a plane, 

Xote in Example 1 that an equation of a is 

3* + 4y - Qs - 6] = T 

that the coefficients of x, y, z in this equation are 3, 4, — 6, that the foot 
of the perpendicular from O to a is {3, 4, —6), that 

32 + 4 s + (_e)* - 61, 

and that 61 is the negative of the constant term in the equation of a. 
Tliis suggests that we transform the given equation of y, 

x - 2y 4 32 - 4 = 0, 

into an equivalent equation in which the sum of the squares of the co- 
efficients is the negative of die constant term, that is, an equation 

ax + hy + cz + d = 

where a 2 4- b' £ 4 e 2 = —d, 

Wc multiply through by k and then determine a positive value of 
k so that the resulting equation of y has the desired property. We get 

fcr - 2ky 4 3Jb - 4fc = 

and we want 

fe3 + ( _2fc)* 4 (3fc) 2 = 4* with fe > 0. 

Therefore 

# + 4/t 2 + 9ft* = 4k, 14*2 = 4^ fc m 2 



12.4 Equations of Planas 529 

Therefore 

Note that 

and that $ is the negative of the constant term of our **adj listed" equa- 
tion of v. 

Taking a clue from Example 1, we suspect that y is the plane that 
is perpendicular at (7,-7,7) to the line through (0,0 4 0) and 
{f t — * I }, and hence that it is the perpendicular bisecting plane of the 
segment with endpoints (0, 0, 0) and (4, — J, 4^ ). To show that it is, we 
let y' be that perpendicular bisecting plane and we show thai y = y> 

Now y ' is the set of all points (st, y, z) each of which is the same dis- 
tance from (0, 0, 0) as it is from (^, — ■$» 4^- ). Therefore 

Y m {{x, y, z) : V* 2 + U 2 + ^ 

= V(* -*) 1 +(* + *)■ + <»- V M. 

/ = ft*. £f, «) : x 2 + if + ~ 2 

Therefore y' = y t and y is a plane. Since 

y = [{*, r/, s) : x - 2y + 3s - 4 = 0}, 
this proves that 

x - 2y + 3z - 4 s 
is an equation of a plane. 

Example 4 Let 5 be the following set: 

5 = {fr y,z):x-2y + 3z = Q). 

Since (0, 0, 0) is an element of 5, it follows that S is a set of points that 
contains the Origin. We shall show that 5 is a plane. 

Taking a clue from Example 2, we suspect that 5 is the plane that is 

perpendicular at the origin to the line through the origin and (1, — 2, 3). 
We suspect further that 8 is the perpendicular bisecting plane of the 
segment with endpoints (I, —2, 3) and (— 1, 2, —3). Let 8' be that 
perpendicular bisecting plane. Wc shall show thai 8' = 5, 



530 Coordinates In Spae* Chapter 12 

We have 

5' = {(x, y, z) : Vfr - l) 2 + {y 4- 2)* 4 (z - 3) 2 

= V(* + I) 1 + (y - 2J 2 + ( Z + 3)*}, 
6" = (fr y t z) : x 2 - 2* + 1 + tp + Ay + 4 + z £ - 6z + 9 

= x 2 + 2r + I + y 2 - 4y + 4 + s 2 + 6s + 9}, 
^ = {(a:, y, z) ; -4* -f 8y - 12z = 0}, 
5' = {{x > y > s):x-2 !/ + 3* = Q}. 

Therefore $* = 8, and 8 tea. plane. 

We arc now ready for several theorems on linear equations in £, y t 
and z, 

THEOREM 12,3 Given an aryz-eoordinate system, every plane 
has a linear equation. 

Proof: Let a plane « !>e given. We consider two cases. 

Case I. O(0, T 0) does not lie on a. 
Case 2. 0(0, 0, 0) lies on <*. 

Proof of Case 1: Let P{a, b t c) be the foot of the perpendicular from 
O to a. Then a, h, c are not all zero. Let Q = {2a, 2b : 2c). Then P is the 
midpoint of UQ, and a: is the perpendicular bisecting pkne of UQ, 

Then 

a = {(*% y, z) : \/x* + y 2 + z 2 

= V(* - 2») 2 4- (y - 2£>} 2 + (z - 2c:) 2 }, 
0£ = {(%, y, z) : ax + by + cz = a 2 + b 2 + c 2 }. 

Therefore in Case I, a has a linear equation. 

Proof of Case 2: Let P{a, b, c) be a point distinct from O on the tine 
perpendicular to a at O. Then a t h, c are not all zero. Let 
Q = {—a. —b, — c). Then O is the midpoint of PQ and q is the per- 
pendicular bisecting plane of FQ. Then 

«= {fcy,*) : Vfr - a) 2 + (y - *>) 2 + {z - c) 2 

= >/(* + «) 2 + fr + *>) 2 + (* + «)"), 

a = {(%, y, *) : x 2 - 2a* + a 2 + y 2 - 2fci/ + fe 2 4- z 2 - 2zc + e 2 
= jc 2 + lax + a 2 4- y 2 + 2by + fe 2 + s 2 + 2sc + c 2 }, 

a m {{x, y, z) : ax 4- by + cz = 0}. 

Therefore iu Case 2, a has a linear equation. 



12.4 Equations of Plum 531 

THEOREM 12.4 Given an ^-coordinate system, the graph of 
every linear equation 

ax 4- by 4- ez + d = 0, 

in which a f b t c,d are real numbers and a., b, c are not all zero, is a 
plane. 

Proof: Let an sy^-coordinate system and an equation 

ax -J- % + cz + rf = 0, 

with a, A», c not all zero, be given. We consider two cases. 

Case J. <*#(). 
Case 2, d = 0. 

Proof of Case 1 ; Suppose d =£ and let 

« = {(x, y, z) : ax + % -f cs + rf = 0}. 

Let /: l)e a number, not zero, to be specified later. Then 

a = {(x, y T z):a f x + b'y -f c'z + d' = 0} : 

where a 1 = fa b* = kh, u' = kc, and ff = hd. Taking a clue from 
Example 3, wc shall determine k t with Jt ^ 0, so that 

d' = -(a'}2 - gtp - {&)*. 
Substituting we get 

kd= -{kaf -{kh)* - (kc) 2 . 
Solving for k we gel 

kd = -Wo 2 - Ffe2 - k 2 c\ 
d = -ka 2 -kb* -kc 2 , 
d = k(-a2 - fea _ ^ 



and this is our specified value of h Note that since a, b t c are not all 
zero, then at least one of the nonnegativc numbers, a 2 , h 2 , c 2 , must be 
positive, and a 2 4- h 2 + c 2 is positive. Hence the expression for k in 
terms of a, b, c, cJ is mathematically acceptable since the indicated di- 
vision is division by a number that is not zero* 
Checking wc find that 

tt = kd = #(-a2 - 68 - c2) = — (Jfcu) 2 - (Mi) 2 - (fee)*, 

d = -(*>)* - (by - { C y. 



532 Coordinates in Space Chapter 1Z 

Let P = (a*, b\ c 1 ) an d Q = (2a' f Zb\ Icf). Let cr' be the perpen- 
dicular bisecting plane of OQ. Then 

or' = {(x, y, 2) : V* 2 + y 2 + £ 

= >/<* - 2a*)* + ( y _ 2bT + > - 2c 1 ) 1 }. 
= {*, i/, s) : *a + y* + & = * 2 - 4rf* 4 4(a')2 

4 y 2 - 4b'y 4 4{fc') 2 + # - 4c'.s + 4(c') a }, 

= (fo y, z) ■ u'x + //./ + <te - («7 - (PT - WT - 0), 
= {(*, y, s) : a'* + fc'y 4 c's 4 tf - 0), 
a {{x, y, z) : kax + kby 4 fcos + kd — $}> 
= {{ar, y, z) : ax 4- by + cz + d = 0}, 

and a' = a. Since & t h a plane, it follows that a is a plane, as we wished 
to prove. 

Proof of Cose 2; Suppose that r / = and let 

a = {{x t y, s) : ox 4 by 4 cs = 0}, 

Then 

a = {(x, y, z) : ox 4 by 4 cz 4 ii = 0}. 

Let P s (a. 6, c)„ p = (— tt, — ft, — c), and let a' be the perpendicular 
bisecting plane of TQ. We shall show that a' = a, and this will prove 
that 

ax + by + cz + d = 

is an equation of a plane. 
Now 

«' = {fa y, z) \ Vfr - tt?» + (y - fr) a + fr - cp 

= \/(x 4 u) 2 4 by 4 b)'* 4 {* 4 c) 2 }, 
a* = {(*, y } *) : a 2 - 2a* 4 0* +• y 2 - 2% 4 P 4 z 2 - ^SJ 4 c 2 
s x' 1 4 2a* 4 a 2 + y t + 2/?y 

4 b 2 4 z 2 4 2zc 4 c 2 }, 
a' = {(x, y, z) : —4ax — 4by — 4cz = 0}, 
ex s {{r, */ t 2) ; ax 4 % 4 ra = 0} , 

"Therefore a' — a. 

THEOREM 12,5 If a, b, c, J are real numbers with a, b, c not all 
zero, then the plane 

a = (£x, y, z) : ax 4 by 4 cs 4 d = 0} 

is perpendicular to the line through 0(0, 0, 0) and P{a, b, c). 



12.4 Equations of Plan** 533 

Proof: If tt = 0, then as in the proof of Case 2 of Theorem 12.4 it 
follows that o is perpendicular to the line Py y where 

Q = (—a, —h, —c) and P = (a, b t c). 

But O is the midpoint of PQ. Therefore Op = P<$ and a 1 OP. 
If d t^ 0, then as in the proof of Case 1 of Theorem 12.4 S it follows 

that a is perpendicular to the line OQ, where 

Q = (2a\2b\2t/)> 
Saw 

0<$ = {{*, tj, z) : x = 2aft, y = 2b% z = toft, t is real}. 
(We have used t as the parameter since k was used for another purpose 

in the proof of Theorem 12.4*) Setting * = -i, we obtain a particular 

2k 

point on OQ, namely (|- , -|- , ^~) = P(a, b t c). Since a is perpendicular 

K ft 1\. 

to Op, and since 5iy = OP, it follows that a is perpendicular to 5?. 



Example 5 Find an equation of the plane perpendicular to OP at F if 
O = (0, 0, 0) and P = (2 4 -7, 0). 

Solution; For every real number d, 2x — ly + d = is an equation 

of a plane perpendicular to OP. (Which theorem or theorems of this 
section are we using here?) To get a satisfactory equation we fix d so 
tlial the plane contains P. Substituting the coordinates of P, we get 

2-2-7(-7) +d = and d = -53, 
Therefore 

2x - 7y - 53 = 
is an equation of the plane perpendicular to Or at P. 

Example ft Find an equation of the plane perpendicular to DP at O 
if = (0, 0, 0) and P = (2, -7, 0), 

Solution: For every real number d, 2x — 7tj + d = is an equation 

of a plane perpendicular to OP. The origin O(0, 0, 0} is a point of this 
plane if2*0 — 7*0 + 4 = 0, that is, if d = 0. Therefore 

2x — 7y = 

is an equation of the plane perpendicular to UP at O. 



534 Coordinate* in Spact Chapter 12 

Example 7 Find an equation of the plane perpendicular to OP at P if 
O = (0, 0, 0) and P = (7, 4, -5). 

Solution: 

7* + 4y - 5z + d = 

7.7 + 4.4 _ 5(_5) + da 

90 + d = 

d a -90 
7* + Ay - 5s - 90 = 

Example 8 Given 

o = {(*, t/, 2) ; 5sc - 2y + 4a - 8 = 0}, 

find u point P .such that « is perpendicular to the line through O(0, 0, 0) 
and P. 

Solution: P - (5, -2, 4). 

Example .9 Given 

a = {(x t y, *) ; 2k - 7* = 0}, 

find a point P such that a is perpendicular to the line through O(0, 0, 0) 
and P. 

SbFtfftbn; F = (2 f 0, —7), 

Example W Find an equation of the plane a that contains the points 
O(0, 0, 0), A(l, 1, 1), and B(l t 7, -3), 

Sofuft'cm; We want real numbers a, b, c y d with a, b, c not all zero 
such that 

a = {(x, y, z) \ ax + by + cz + d = 0). 

Since O, A, B arc points of a, it must he true that 

(1) a*O + ft*O + 0*O + <la>0, 

(2) a •! + b'l + cl + d = (), 

(3) a • I + b • 7 + &•{— 3) + 4 = a 

From (1) we deduce that d = 0. From (2) and (3) we deduce by sub- 
traction that 

1 -6/;-f4c = 0. 



12.4 Equations of Plan** 535 

Now neither b nor c can be zero. Why? To complete the solution we 
take any number except for b and solve for the corresponding a and c. 
Thus, if b = — 2, then c = — 3, a = 5, and 

a a {(x, y, z) : 5x - 2y - 3* = 0}. 

To check, substitute coordinates as follows: 

O: 5*0-2-0-3-0 = 0-0-0 = 0. 
A: 5-1-2-1-3-1 = 5-2-3 = 0. 
B: 5*1-2-7-3- (-3) = 5 - 14 + 9 = 0. 

Kxamplc II Find an equation of the plane that contains the three 
points P(l, 5, 7), Q(-l, 2, -4), R% 1, -5). 

SoJti (ton; 

(1) ax + % + ez + d = 

(2) a • 1 + /? - 5 + c • 7 + a" = 0, a + 5b -f 1c + d = 

(3) fl-(-l) + ft-2 + c-{-4) + d = 0, -a + 2fo-4e + d = 

(4) a -2 + fc * 1 + c - (-5) + 4 = 0.2<t + &-5c+<Z=0 

From (2) and (3) we get (5), and from (3) and (4) we get (6). 

(">;> 2a + 3& + lie = 

(6) -Za f b + c = 

From (6) we get (7), and from (7) and (5) wc get (8). 

(7) -9a + 3b + 3c = 

(8) llo + 8c = 

Take a = 8; then c = — 11. From (6) we get fc = 35 and from (2) we 
get d = — 106, A satisfactory answer is 

8* + 35y - llz - 106 = 0, 
Check: 



P: 8-14-35-5-11*7- 

= 8 + 175 - 77 -- 106 = 0. 

Q: 8-(-l) + 35-2 - 11 -(-4) - 106 

= -8 + 70 + 44 - 106 = 0. 

R: S-2 + 35-1 - ll-(-5)- 108 

= 16 + 35 + 55 - 106 = 0. 



536 Coordinates In Span Chapter 12 

THEOREM 12.& Consider the plane 

a ={(*» if* x ) : ax + % + <* + d - °)' 

If « = 0, a is parallel to the x-axis. If h = 0, a is parallel to the 
y-axis. If c = 0, or is paralU-] to the --axis. 

ftoo/v Suppose first that a = and d = 0. Then 

a = {(*, y t z):by + cz = 0} 

and for every real number x, a contains the point (jr., 0, 0). Therefore of 
contains die ar-axis and is parallel to it. 

Suppose next that a = and fi 7^ 0. Then 



and since 



a = {{x, y, z) : by + cz 4- d = 0} 



b-O + c-O + d^O, 



it follows that no point (x, 0, 0) is a point of a and henec that the x-axis 
is parallel to id 

Similarly, it may be shown that if b = 0, men a is parallel to the 
ff-axis, and that if C = 0, then a is parallel to the s-axis. 

Examph 12 Let a, /?, y, 5 be given as follows: 

a = {(*, y t z) : 3* + 4y - 12 = 0}, 
/?={(*, i,,z):3x-2z = 0}, 
7 = {(x, tj, z) : 5tj = 8}, 
* = {(*, i/, «) : 2* - 5 = 0}. 

(See Figure 12-7,) & is parallel to the s-axis, a contains the line I in 
the xy-planc widi x- and (/-intercepts 4 and 3, respectively, and every 
line parallel to the s-axis that passes through a point of I 




Figure li-7 



12.4 Equations of Plane* 937 

(See Figure 12-8.) /? is parallel to the y-axis; in fact, it contains the 
i/-axis. The plane /? contains the line I in the .re-plane that passes 
through the points ((I } 0) and (2, 0, 3). It contains every line parallel 
to the y-axis that passes through a point o( L 




Figure 12-8 

(See Figure 12-9.) y is a plane parallel to the ar-axis and the 3-axis; 
hence it is parallel to the .ts-plane. y contains every point whose (/-coor- 
dinate is 1.6 and every point of y has y-coordinale 1.6. 



♦a 




yml(x,y,i};5y>B\ 



Fignn- 12-W 

(iSee Figure 12-10.) 5 is a plane parallel to the y-axis and to the 
>axis; hence it is parallel to the y£-plane. The plane 5 is the set of all 
points whose ^-coordinate is —2.5, 




*- -if*, y.*): 2* + *-0f 

Figure 12-10 



538 Coordinates in Space Cnap:e'12 

EXERCISES 12,4 

■ Let a plane « = {(*, y, s) : 4x — 3y + 180 = 0} be given. In Exercises 
1-10, determine if the given point is a point of a, 

1. (0, 90, 0) 6. (0, W, 137) 

2. (45.0,0) 7, (-45, 0,237) 

3. (0,0,180) 8, (-30.20,10) 

4. (0, 0, 0) 9, (7, 69, j) 

5. (0, 0, 37) 10, (-100, -73± 6Xif) 

B In Exercises 1 1 -20, an equation of 8 plane is given. Find three numbers 
Or b, c f not all zero, such that the given plane is perpendicular to the line 
through the origin and P{a, b, c). 

11. -3.r - 7y + 14 = 16, f-r ? -nl=0 

12. 2x - y + z - 73 = 17. x - y + 10s = 

13. 2y = 3x 18. x - y + 10s + 10 = 

14. tj = 3x + 5 19. y + lOx - 4s = 

15. * = 3y — 3 20. x - y = s - 1 

■ In Exercises 21-25, siin plify the given equation to an equivalent equation in 
the form of the general linear equation ax + by + ex -+■ d = 0. If one of 
the coefficients is zero, Uiat term may be omitted in the simplified form. 
Thus we accept 3* + 4y = in place of 3* + 4tj + Oz + = 0. 

21. yx" + tf + z* = V(x -l )»f (y-2)' + (g- 3) s 

22. y '{* - 2}* + (y - 3?' + s' = \ /^T 2f+(y + 3)' + ^ 

23. V fx- I) 2 + (r/ - if ^UP e V (x- 12j a + ( y - 5) a -f s* 
M, y ^T4) g + (y + 4)» + (a -T ffi = V* 2 ^ y» + ^ 

25. Vfx^T? + ~y*" + [5 +TP = y/{x - 1)* + (y - 1)2 + (z + 1)2 

■ In Exercises 26-30, find a linear equation in simplified form for the plane 
that con tarns die diree given points. 

26. (0, 0, 0), (8, a 0), (-3, 0, 1) 29. (8, 0, -4), (0, 4, 3), (4, 4, 1) 
21 (1, 3. 1), (0, 1, 1), (2, 1, 1) 30. (3, 0, 1), (-9, 0, -3), 

23, (5, 2, 3), (-3. 6. 7), (0, 4, 6) (51, 10, 17) 

B In Exercises 31-40, ail equation of a plane is given. Find the coordinates 
of its intercepts on the coordinate axes. If it does not intersect an axis, write 
"none." The answer to Exercise 31 is given as a sample. 

31. 2x - 3s + 30 = x-ratercept; (-15, 0, 0) 

{/-intercept: None 
s- intercept: (0. 0, 10) 



12,5 Symmetric Equations tor a Un« 539 

32. 3x + by - 2x - 20 = 37. tj = -3 

33. * - y + s = 38. z = -4 

34. 5* - 4y + 100 = 39.x + y -J- e = 100 

35. 2x - ij + -s = 40. a : 4* y + = + 100 = 

36. x = 6 

In Exercises 41-50, state whether ihc given set S is a plane, the union of two 
distinct planes, n line, the union of two distinct lines, a set consisting of a 
single point, or the null set. 

41. S = {(x, y t z) : x = 2, y = 3, z a 4} 

42. S = {(x, y. a) : x + y + s = 5J 

43. S={{x t ,j,z):x+,j = 5} 
44.S={(x t y):x+y = 5) 

45. S = {(», y) : x = 5} 

46. S = ((x, y, z):x = 5orx= 10} 

47. S = {(x t y, x) : x + p + « = 10 and x + y + z = 20} 
45. S = {(x, y, x) : x + y -h z= 10 or x 4- y -f s = 20} 

49. S = {(x, y, z) : x + y + 2* = 10 and x - 2y + 2z = 10} 
30. S = {(x, i/, c) : x + y + 2s = 10, x - y + 2* = 10, 2 = 0} 



12.5 SYMMETRIC EQUATIONS FOR A LINE 

Since a linear equation in x and y represents a line in an xry-plane 
and a linear equation in x, y> z represents a plans in xj/z-spacc, it is nat- 
ural lo ask how we might represent a line in terms of coordinates in 
xys-spaee and to do it without using a parameter. A clue to the answer 
is provided by some of the exercises of Exercises 12.4. As usual, we 
assume an xyx-eoordinate system is given. 

Example 1 If P = (2, l s 7) and Q = (5, -1, -3), then 

The notation 

x- 2 y- 1 x- 7 

5-2~-l-l : -3-7 

is a short way of saying that 

ill, y 1 and y 1 = *- 7 . 

5_2 -1-1 -1-1 -3-7 



540 Coordinates in Space Chapter 12 

To see that the representation in terms of these equations is correct, 
note that 

* - 2 - y - J rt „i j/- 1 _ ~- 7 



unci 



5-2 -1-1 -1 - 1 -.3-7 

arc equations of distinct planes, one of them parallel to the s-axis and 
the other parallel to the it-axis. Note that both F and Q lie on each of 

these planes. (Check this by substituting coordinates.) Hence FQ is 
precisely the intersection of these two planes and the representation 
we gave in terms of "symmetric" equations is a satisfactoiy one. Actu- 
ally, there is more symmetry in this situation than we have indicated 

so far. It is natural to think of ry as die intersection of three planes 
related to the three expressions equated to each other in the set-builder 

notation, Thus every point of FQ lies on each of the following three 
planes, a t /?, and f, which are parallel, respectively, to the a>, y-, and 
2>axes: 

* = [<* * *> ■■ fff = -^nrr)- 

Generali zing the situation in Example 1, we obtain the following 
theorem. 

THEOREM 12. 7 If P(x lt y it *i) and Qfa y 2i z 2 ) are two distinct 
points with x% ^= 4*2, tj\ ^= tj>£, £\ ^ jgtg, then 

FQ = { lx t y, z) : = -* *— = - 

I X2 - Xl yz - Jfr 8£ - «i 

ftw/: Let 

I »2 — 3^1 ^2 — 2»i ) 

I x 2 — xi y 2 — yi J 



12.5 Symmetric Equations for a Lin* 541 
Now P(xu j/i, 2i) is a point of a since — ^- = — — = 

y* — yi z z ~~ *h 

and Q(x 2 , ijt, %) is a point of a since - /a " yi = * 2 ~ Zl = 1. 

^a - 1/1 «■— »i 

Similarly, we can show that P and p each lie on both /3 and y , 

Also, a, /i, y are three distinct planes. For a contains the point 
(xi — X t tf], Zi} t whereas ft and 7 do not; fi contains (xi, 1/1 — 1, Zi), 
whereas on and y do not: y contains (xi, yi,Z\ — 1), whereas a and fi do 
not The — 1 here may appear to border on the magic. Actually it does 
not For example, if Xo is any nonzero number whatsoever, then a con- 
tains (xi — Xo t yi, zi), whereas fi and y do not 

Therefore (x, y, z) is a point of PQ if and only if it is a point of all 
three of the planes a, fi, y, and hence the representation using sym- 
metric equations is a valid one. 

THEOREM 12.8 If F(x u 1/1, *t) and Q(x 2 , y 2 , $§} arc distinct 
points with x-j ^= x 2 and y^ sfi y 2 > then 

I X 2 - Xl $2- IJl > 

Proof; 

X — Xi u — 5/1 , 

and z = «i 



ia - xi ya - ^3 

are equations of two distinct planes each containing both P and Q, 

Therefore PQ is the intersection of two planes, and the representation 
given in the statement of the theorem is a valid one. 

Similar proofs may be given for Theorems 12*9 and 12,10. 

THEOREM 12.9 If F(x u yu z t ) and p(x 2 , yu z^} are distinct 
points with xi ^= x 2 and z± =£ ffg, then 



i v*> y> « 



X — Xi * — Zl 

'} 



Zi \ 

X 2 — X| Z% — Zi i 

1 HEO /t EM 12.10 Tf F(Xi , «/ 1, ?i ) and ()(xi , i/ 2 , z 2 ) are two distinc I 
points with tji ffe 1/2 and zi 7^2. then 

I y-i - yi *2 - zi ) 



542 Coordinates. In Space Chapter 12 

Proofs of Theorems 12,11, 12.12, 12.13 arts assigned as exercises. 

THEOREM 12 .1 1 If P(x lt y h ft) and Q(xi, y lt * 3 ) are two distinct 
points, then 

W = ((*. y.z}:x = xi and y = y t }. 

THEOREM 12. 12 If P(x lt y x > zi) and Q{x x , y 2t Si) are two distinct 
points, then 

¥@ = {{*, y t z):x = xi and *= %). 

THEOREM 12.13 If flfa, ^i> *i) and 9(* 2 , if! . a,) are two distinct 
points, then 

PQ = {(*, y, s) : y = yi and s = *i). 

EXERCISES 12,5 

■ In Exercises 1-5, a tine is given in terms of symmetric equations. In eac-h 
case, find the missing coordinates of P t Q t R so thai they will be points of the 
given line. 

m. ib m x e& m. m ). «{3. in. h ) 
"£3. [Em). «7. a, >, j!(n, s h ) 

1 -5 3 2 J 

nm m, -t), p( m, m, -3), a® m, 1 1 
fc(fc*4>f-f-i}- 



12.5 Symmetric Equation* for a Line 543 

■ In Exercises 6-1 0, a line is given. To which of the coordinate axes, if any, is 
it parallel? 

10. fa y, z) : x = 3, y = 4} 

■ In Exercises 1 1— IS, a Hue is given in terms of parametric equations. Repre- 
sent the line using symmetric equations. 

11. {(,*, y,z): x = 1 -f 3k, r/ = 2 — 2k, z = 3 — 4Ar„ fc is real} 

12. {{x, y,z):x = - 1 - k, y = 2 + 4k, z = 5 - k, k is real} 

13. ((x. {/, z) : .t = A\ y s 2k t * = 3fc, fc is real} 

14. {(x, ^ z) : x = 3 - 4k y y = 2fc, 2 = 1 - k, k is real) 

15. {x, y. z) : x = 2 + ft(5 - 2), y = 3 + Jk{5 - 3), 2 9 2 + fc(l - 2), 
k is real} 

■ In Exercises 16-20, a plane and a line are given. Find their point of 
intersection. 

lfi. ((x, i Jt z) : 3x + 4y + 55 = 80}, {(*, I/, jj : -|= -| « -|) 

17. {{x, j/, 5) : 2x -1- y - 10 = 0} T {(x T r/, s) ; * = 5 and 3 = 3} 

18. {{x, ,j,z):x-y + z = 0}, [{x, y, z) ; ^1 = ±^A = i^.} 

16. {(x, y, z) : 2x - y + z + 5 = 0}, 

20. {{x, j/,s) : x + «/ + 3 + 1 = 0), {{x, $f. 3) : % = y = z) 

21. Prove Theorem 12.9, 

22. Prove Theorem 12,10. 

23. Prove Theorem 12.11. 

24. Prove Theorem 12.12. 

25. Prove Theorem 12.13. 



544 Coordinates in Space Chapter 12 

CHAPTER SUMMARY 

We began this chapter by defining an xj/jb-COORDINATE SYSTEM. 
Stalling with a pair of perpendicular lines called the x- and the y-axcs, we 
know that there is one and only one line through their intersection that is 
perpendicular to both of them* We call this line the z-AXIS, The one and 
only point that lies on all three of the coordinate axes is called the ORIGIN'. 
There are three coordinate axes and three coordinate planes. Each axis lies 
in two of the coordinate planes and is perpendicular to the oilier one. All 
lines perpendicular to a coordinate plane are parallel to one of the coor- 
dinate axes. AH lines parallel to a coordinate axis are perpendicular to one 
and the same coordinate plane. If a plane is parallel to a coordinate plane, 
then it is perpendicular to a coordinate axis and to all lines parallel to that 
axis. 

In our work with coordinates in space we stated and proved the 
DISTANCE FORMULA THEOREM and slated a theorem regarding 
parametric equations for a line, These theorems are natural extensions of 
some theorems for the two-dimensional case in Chapter 11, The Distance 
Formula and parametric equations for a line are useful tools in solving prob- 
lems involving solid geometry. Linear equations may be used to represent 
planes, and combinations of them in what we call symmetric form may be 
used to represent lines. 



REVIEW EXERCISES 

In Exercises 1-8, eight points are given as follows: A(0, 0, 0), 13(3, 0, 0), 
C{3, 4, 0), D(Q, 4, 0), £(0, 0, 2), F(3, 0, 2), G(3, 4, 2), H(0, 4, 2). 

1. Draw a graph of the rectangular solid with vertices A. B, C, D, E. F, 
G,H. 

2. Prove that lABF^ L DBF — / CBF =z LEFC. 

3. Find the measure of dihedral angle II-EA-F, 

4. Prove that AG = EC ss W =s ¥D. 

5. Find the distance from point A to plane CDH. 
& Find the distance from point A to plane FOIL 

7. Find the distance hetwecQ plane ABC and plane EFC2. 

8. l^et I. J, K be the midpoints of EE, ItfZ, DC ( respectively. Find the dis- 
tance between the planes JJK and ECC. 

In Exercises 9-13, find the distance between the two given points, 

9. (Xi, t flt %) and (jfifc t/ 2 , z*). 12. (7, 8, 0) and (10, 12 t 0). 

10, (a, b, c) and (d, e, /). 13. (3, 4, 5} and (8, 9, 10). 

11. (p, q, r) and (p t q, t). 



Review Extrclsts 545 
In Exercises 14-22, there are two given points: A{1, 2, 3) and #(4. 5, 6)* 

14. Express AB using parametric equations and set-builder notation. 

15. Express AB using parametric equations and set-builder notation. 

16. Express BA using parametric equations and set-builder notation, 

17. Express opp AH using parametric equations and set-builder notation. 

18. Express opp BA using parametric equations and set-builder notation. 
10. Find C on AB if AC = 10 - AB. 

20. Kind C on opp AS it AC = 10 * AB. 

21. Find the bisection points of AE. 

22. Find the two points winch divide AB internally and externally in the 
ratio ,. 

23. Given A(5, 1.1). fl{3, 1, 0), C<4. 3, -2). JJ(6 t 3 T -1}, prove that quad- 
rilateral ABCD is a parallelogram. 

24. Given A(2, 4, 1), B(l, 2, -2), C(5> 0, -2), pro%-e that AABCis a right 

triangle, 

25. Find three noncollincar points on the plane given by the equation 

3x + Ay - z + 15 = 0. 

26. Find three distinct points on the line given by the symmetric equation 

x—2 _ y-4 _ z+3 
3 ~ 2 5 ' 

27. Write an equation of the plane that is perpendicular at (2, 4, — 3) to the 
line of Exercise 26. 

28. Find the x-, y-, and s-intercepts of the plane a where 

a = {(jr, tj t s) : 3x - 7y - fa + 5 = 0}. 

29. Express the line / where 

I = {(*, sj, z) : x = 1 + 2*, y = 1 - 3fc, z = k,k is reu]}, 

using symmetric equations but no parameter. 
3(1. Consider the line I and the plane or perpendicular to / at C>(0, 0, 0), given 
as follows: 

Find two distinct points P and Q on J such that « is the perpendicular 
bisecting plane of T$, 




/ 



\ 




4 







Chapter 




Erick ttartmoimf Magnum Photot 



Circles 
and Spheres 



13.1 INTRODUCTION 

An the title suggests, this chapter is concerned with properties of 
circles and spheres, some of which you may already have studied in 
your earlier work in mathematics. 

The first part of the chapter deals with the intersection proj)erties 
of a circle and a line in the plane of the circle and the intersection 
properties of a sphere and a plane. 

The second part of the chapter is about the degree measure of arcs 
of a circle and properties of certain angles in relation to arcs, secants, 
tangents, and chords. We also consider properties of lengths of secant- 
segments, tangent-segments, and chords. 

13.2 CIRCLES AND SPHERES: BASIC DEFINITIONS 

Up to now, in our formal geometry, we have not discussed "curved 
figures/' that is, figures not made up of segments, rays, or lines* Perhaps 
the simplest of these figures are the circle, the sphere, and portions of 
a circle, called arc*;. We begin with some formal definitions. 



548 Circles and Spheres 



Chapter 1 3 



Definition 13.1 (See Figure 13-1.) Let r be a positive num- 
ber and let O be a point in a given plane. The set of all points 
Fin the given plane such that OP — r is called a circle. The 
given point O is called the center of the circle, the given 
number r is called the radius of the circle, and the number 
2r is called the diameter of the circle. 



A circle is a curve and encloses a portion of a plane. As an illustra- 
tion, consider a circular disk. The edge of the disk is what we have in 
mind when we think of a circle. The edge of the disk together with its 
interior points is what we have in mind when we think of a circular 
region. Wc will have more to say about a circular region (that is, a circle 
and its interior) in Chapter 14, 

Question: Is the center of a circle a point of the circle? Explain, 





Circle with center O 
and radius OF = r 



^V 



Sphere with center 

and radius, OP = r 



Figure 13-1 



Definition 13.2 (See Fij^ire 13-1.) Let r be a positive Dum- 
ber and lei () be a point in space. The set of all points P in 
space such that OP = r is called a sphere. The given point O 
is called the center of the sphere, the given number r is 
called the radius of the sphere, and the number 2r is called 
the diameter of the sphere. 



A sphere is a surface and encloses a portion of space. As an illus- 
tration, consider a ball. The surface of the ball is what we have in mind 
when w-e think of a sphere. The surface of the ball together with its 
interior points is what we have in mind when we think of a spherical 
region. We will have more to say atout a spherical region (that is, a 
sphere and its interior) in Chapter 15, 



13.2 Circles and Spheres: Banc Definitions 549 



Definition 13.3 Two or more coplanar circles, or two or 
more spheres, with the same center are .said to be concentric. 



In Figure 13-2, Q is the common center of three concentric circles 
with radii (plural of radius) r-i, r 2l and r 3 „ respectively. 




Figure 13-2 

Let O be a point in a given plane a. Then we can choose an xy-coor- 
dinatc system in a such that the origin is at O. Let Cbe a circle in a with 
center at O and radius r. Let P(x, y) be any point on C as shown in 
Figure 13-3. 



Pto, y) 




Simon 13-3 



Then, by the Distance Formula, OP = r = >/(x - 0)- + (y - G) 2 . 
Therefore r = \/^~+ y 2 and x 2 + y- = r 2 . 

Converse ly, if F(x, y) is any point such that x 2 + y* = f*, then 
y/( x - 0) 2 + (y - 0) 2 = r T OF = r, and F is a point of the circle C. 
Thus, in a given xt/-plane, the circle C with center at the origin and 
with radius r is given by C = {{x, y) ; x 2 + y 2 = r 2 }. 

We have proved the following theorem. 



550 Circles and Spheres. 



Chapter 13 



THEOREM 13.1 Let an xy-phme be given and let O be the origin 
and let r be a positive number. Let C be the circle in the sy-plane 
with center O and radius f. Then 

C = {(*, y) : ** + r/2 = r*}. 



Definition 13.4 A chord of a circle or a sphere is a segment 
whose endpoints are points of the circle or sphere. A secant 
of a circle or sphere is a line containing a chord of the circle 
or sphere- A diameter of a circle or sphere is a chord contain- 
ing the center of the circle or sphere. A radius of a circle or 
sphere is a segment with one endpoint at the center and the 
other endpoint on the circle or sphere. 



In Figure 13-4, C is a circle with center P, and S is a sphere with 

cent er P. For the circle and the sphere, AB is a chord, 46 is a secant, 
ED is a diameter (and also a chord), and PQ is a radius. The endpoint of 
a radius that is on a circle or a sphere (such as point Q in Figure 13-4) 

is often referred to as the outer end of that radius. Is ED in Figure 13-4 
a secant? It follows from Definition 13,4 that a secant is a line that in- 
tersects a circle (or sphere) in two distinct points. 



*<* 





Figure 13-* 

Note that, in connection with a circle or a sphere, the word "radius" 
is used in two different ways and the word "diameter" is used in two 
different ways: (1) each word 1$ used to mean a certain segment and 
(2) each word is used to mean the positive number that is the length 
of a segment. Hus should not be confusing because the context in 
which the word is used should make it easy for you to decide which 
meaning Ls intended. For example, if we speak of a radius or a diameter 
of a circle or of a sphere, we mean a segment. If we speak of the radius 



13.2 Circles and Spheres: Basic Definitions 551 

or the diameter, we in can the number that is the length of a segment. 
Thus a circle has infinitely many different radii if radius is interpreted 
to mean a segment; it has just one radius if radius means the length of a 
segment. 

Just as we speak of congruent angles, or congruent segments, or 
congruent triangles, we often speak about congruent circles or con- 
gruent spheres. How would you define congruent circles or congruent 
spheres? Does your definition agree with the following one? 



Definition 13,$ Two circles (distinct or not) are congruent 
if their radii arc equal. Two spheres (distinct or not) are con- 
gruent if their radii are equal. 



Using Definition 13.5, it is not hard to prove that congruence for 
circles (or spheres) is an equivalence relation; that is, it is reflexive, sym- 
metric, and transitive. Thus if A, B, C are any three circles (or spheres), 
it is true that 



I A=A. 

2. IfA^B, thcnBsA. 

3. Tf A == B and B s C, then AsC. 



Reflexive Property 
Symmetric Property 
Transitive Property 



EXERCISES 13.2 



In Exercises 1 10, refer to the circle with center F shown in Figure 13-5. 
Assume that all the points named in the figure are where they appear to l>e 
in the plane of the circle. Copy and replace the question marks with words 
or symbols that best name or describe the indicated parts. 

1. TJE is called a [7] of the circle. 

2. FE is called a [T] of the circle. 

3. AH is called a [Tj of the circle. 
AB could also be culled a (T] 
of the circle. 

4* FG is called a (7] of the circle. 
S. FG is called a (7] of the circle. 
& DE is called a [7] of the circle. 

7. C is the outer end of the radius [TJ* 

8. A is the |7J of the radius |7]. 

9. The points named in the figure thai arc points of the circle (that is, on 
the circle) are |7], 

10. The points named in the figure that arc not points of the circle are [?]■ 




l-i^m. U-.- 



552 Circles and Spheres 



Chapter 13 



Li Exercises 11-15, refer to the Sphere with center Q shown in Figure 13-6. 
Copy and replace the question murks with words or symbols that best name 
or describe the indicated parts. 




Figure 13-9 

11. QR is a [Tj of the sphere. 

12. If S, Q, T are collincar, then ST is a [TJ of the sphere, 

13. KS is a \?} of the sphere, 

14. RT h a [?] of the sphere. 

15. Points T| are outer ends of given radii. 

16. Prove that if two circles are congruent, then a diameter of one is con- 
gruent to a diameter of the other. 

17. Prove that congruence of circles is an equivalence relation, (You will 
need to use Definition 13.5 and tbe theorem that congruence of seg- 
ments (radii) is an equivalence relation.) 

In Exercises 18-25, refer to the circle with center at the origin of Lhc xy- 
conrdinatc system shown in Figure 13-7, P{x t o) is a point on the circle. 



Figure 13*7 



! 








< 


^y 












































































j* 


















P( 


"K 






















/ - 


V * 








{ 








\* 


q 


-6 1-4 .3.3-1 t 


12 3 4 


1 6 ' 


\ 


-- 


1 








/ 






































































1 


1 















13.2 Circles and Spheres: Basic Definitions 553 

18. Use the Distance Formula and express the distance between O and P 
in terms of x and y. 

19. Write an equation like that in Theorem 13. 1 of the circle with center 
at the origin and radius 5. 

20. Write the coordinates of the points on the x- and y-axes that arc on the 
circle of Exercise 19. Check your answers by substituting the coordi- 
nates in the equation of the circle. 

2L Is A = (3, i) u point of the circle of Exercise 19? 

22. Is B = ( — 4, 3) a point of the circle of Exercise 19? 

23. Is K = (l t 2%/^) a point of the circle of Exercise 19? 

24. Is R = (—2, 4.5) a point of the circle of Exercise 19? 

25. Write the coordinates of two more points (different from those in Ex- 
erases 20-24) that lie on the circle of Exercise 19. 

26. Given the ,t [/-coordinate system shown in the figure with A = (1,2), 
P b (x, y) y and AP — 5, use the Distance Formula and express the dis- 
tance between A and Fin terms of x and y, 







' 


'>' 






, 












. 




4 «*.*j 




4 




*\ 












4(1.2) 























x * 




-2-1 


12 3 4 


s a 






■ 










i 











27. In Exercise 26, write an equation like that in Theorem 13.1 of the circle 
with ecntcr at A and radius AP = 5. Is the point S = (4, 6) a point of 
this circle? Write the coordinates of at least three more points that are 
on this circle. 

25. Which of the following are points of C as {{%, y) : x 2 + y 2 = 100}? 
(a) (a 10) (b) (-6,8) (c) (4,9) (d) (2^ -2\/I5) 

29. Whet is the nidi us of the circle of Exercise 28? 

30. Find five more points of C in Exercise 28. 

Exercises 31-34 concern the set C, where C - {(x, y) : i* + y 2 = 16). 

31. Is C a circle? Why? 

32. Find x if {x, 3) is a point of C. (There are two possible values for a ; 

33. Find y if (4. y) is a point of C. 

34. Can you find x so that (x, 5) is a point of C? Explain. 



554 Circles and Spheres Chapter 13 

■ Exercises 35-37 concern the circle C» where C = {(x, y) : x 2 + y z = 36). 

35. What restrictions on x and t/ would give only the part of the circle in 
quadrant 1? 

36. What part of the circle would you be considering under the restriction 
x 2 + y 2 = 36 and x < 0? 

37. What restrictions ou x and y would give the intersection of C and quad- 
rant 111? 

38. challenge problem. Let an xi/-plane with A = {h, k), F = (x t y), 

AP = r > be given. Write an equation in terms of x, y, h f k t r for the 
circle with center al A and radius AP = r. 



13,3 TANGENT LINES 

If you look at a drawing of a circle in a given plane, it is easy to see 
that the circle separates the points of the plane not on the circle into 
two sets. One of these sets consists of those points of the given plane 
that are "inside" the circle and the other set consists of those points 
of the given plane that are "outside" the circle. 



Definition 13.6 (See Figure 13-8.) Let a circle with center 
O and radius r in plane a he given. The interior of the circle 
is the set of all points P in plane a such that OP < r. The 
exterior of the circle is the set of all points P in plane « such 
that OP > r. 




Home 13-S 




It is clear from Definitions 13,1 and 13.6 that if F is anv point in 
the plane of a circle with center O and radius r, then P is on the circle 
{OP = r), or P is in the interior of the circle (OF <[ r), or P is in the ex- 



13.3 iMpii Lines 555 



terior of the circle {OP > r). Wc sometimes say "Pis inside the circle' 
or "P is outside the circle*' when we mean "P is in the interior of the 
cirde" or "Pis in the exterior of the circle. 1 ' respectively. 

Figure J 3-9 shows an ^-coordinate system whose origin is the 
center of a given circle C with radius r. The figure also includes expres- 
sions for C, its interior and its exterior, in terms of coordinates using 
set- Wilder notation. 



Inferior of C =• {(*; Jf) 
F.xlerk* at C = {(*»jr) 



«» + 9* = »*) 

•* + V* > **) 




Figure 13-9 

Definition 13.7 If a line in the plane of a circle intersects 
the circle in exactly one point, the line is called a tangent to 
the circle and the point is called the point of tangeney,or 
die point of contact. Wc say that the line and the circle arc 
tangent at this point. J f a segment or a ray intersects a circle 
and if the line that contains that segment or ray is tangent to 
the circle, then the segment or ray is said to lx* tangent to 
the circle. 



In Figure 13-10, if Ms a line in the plane of circle C and if the in- 
tersection of I and C Is just the one point P, then I is tangent to C at P. 




Figure 13-10 



5 56 Circles and Spheres 



Chapter 13 



Given a circle and a line in the plane of the circle, what are the possi- 
bilities with regard to the intersection of the line and the circle? Figure 
13-1 1 suggests that there arc only these three possibilities: the inter- 
section of the line and the circle may be the empty set, or it may be a 
set consisting of exactly one point, or it may be a set consisting of 
exactly two points. 



cnl=\P\ t 




blgwo 13-11 



That these arc the only three possibilities can be verified in the fol- 
lowing way. Suppose that wc are given a circle C with center in plane 
a and that P is a point in plane a. Then P is outside the circle, as shown 
in Figure 13-1 la, or P is on the circle, as shown in (b) t or P is inside 
the circle as shown in (e). 

If Pis outside the circle, we shall show that the unique line I in plane 
cr such that OP _^ I at P does not intersect the circle. If P is on the cir- 
cle, wc shall show that the unique line / in plane a such that OP ± I at 
F intersects the circle in exactly one point (hence / is tangent lo C at P). 
If f is inside the circle, then either P = O or P=£CK1( P / 0, we 
will show that the unique line I in plane a such that OP I I at P inter- 
sects die circle in exactly two points which are equidistant from P, 
Finally, if P = O, we will show that any line I which contains P inter- 
sects the circle in exactly two points. We are now ready for the follow- 
ing theorem. 



33.3 Tangent Lines 557 

THEOREM 13.2 Given a line I and a circle C in the same plane, 
let O be the center of the circle and let P l>e the foot of the per- 
pendicular from O to line I, 
L Every point of 2 is outside C if and only if P is outside C 

2. I is tangent to C if and only if P is on C 

3. I is a secant of C if and only if P is inside C. 

Proof: Let r be the radius of C and let OP = a. We select an ^-coor- 
dinate system in the plane of C and I with the origin at O s with the 
y-axis parallel to I and with P on the nomiegative ar-asis. Then 



and 



P=(a,0)> 
Z= {(*,*/) :* = a}. 



Proo/ 0/ i; Suppose we are given that ? is outside C as shown in Fig- 
ure 13*12; then a > r]> 0. Why? It follows that a 2 > r 2 and hence 
Q 2 + y 2 > **• Therefore all points (fl* y) are outside C, Since 

t-{{x i y):x=a} = {(a, y) : y is real}, 
it follows that all points of J ate outside C. 




figure 13-12 



Now suppose that I is outside C; then every point of I, including P % 
is outside C and the proof of (1) is complete. 



558 Circles and Sphert* 



::>spter ".3 



Proof of 2: Suppose wc are given I hat F is on C as shown in Figure 
13-13; then a = r. Why? The intersection of C and I is 

Ifa y) : x* -f i/ = r*} fl {(*, y):x = u} = {(a. y):y* = Q}> 



c 





Ij 






Pte.D) 



FlRUIT 13-13 



Since zero is the only number whose square is zero, it follows that 
y = 0. Therefore the only point of intersection of I and C is P{m 0). 
Thereforc I is tangent to C at P. Why? This proves that I is tangent to 
C if P is on C, 

Now suppose it is given that I is tangent to C. Then J and C have 
exactly one point in common. This means that I H Cis a set consisting 
of exactly one point. But 

C = {{*, y) : x* + f = i*}, 

and 

IDC= {{x, \j) : x* 4- J/ 2 = r 2 and* = a} 
-{fey):* 4 E/ 2 = r2}. 

If (a, tj) with t/ ^ Q is a point of Z fl C, then (a, -y) is also a point of 
I H C and there are two distinct points in I D C. Since there is only 
one point in / fl C, it follows that y = and a z = #, Since a > 3 it 
follows that a m f and the one and only point in I H C is the point 
HA 0)* This proves that / is tangent to C only if P is on C, 



13.3 Tangent Un« 559 

Proof of 3: Suppose w© are given that P is inside C as shown in Figure 
13-14a or b. Recall that O is the center of the given circle C Then 

P = («, 0) where < a < r> 



an< 



I n G = {(*, y) : x = a and * 2 + t/ 2 = r 2 } 

Since < a < r, il follows that 

r* - o 2 > 0, y/fi-d* ^ - Vr^fl* 

and the points (a, "v 71,2 — ° 2 ) an ° 1 (*>■ — V 4 — ° 2 ) are distinct points. 
This proves that / is a secant of C if P is inside C. 




Suppose now that I is a secant of C; then / intersects C in two dis- 
tinct points which we have just shown to be (a. yV 2 — a 2 ) and 
{a> — y/f& — « a ). This means that r 2 > « a . (Why is it that we cannot 
have r a = a 2 or r 2 < a 2 ?} Since r > and « > 0, it follows that r > a 
and hence that OP <C ft Therefore F(a, 0) is inside C and the proof is 
complete. 

Now that we have proved Theorem 13.2 we proceed to state our 
first basic theorems on tangents and chords. To prove some of these 
theorems wc need refer only to Theorem 13.2 and see which of parts 
(1), (2), or (3) apply. 



560 Circles and Spheres 



Chapter 13 




THEOREM 13.3 Given a circle and a line in the same plane, if 
tlie line is tangent to the circle, then it is perpendicular to the radius 
whose outer end is the point of tangency. 

Proof: Let C be the given circle with center O, let line I be tangent 
to C at R, and let P be the foot of the perpendicular from O to /. it fol* 
lows from part 2 of Theorem 13.2 that P is on C. (See Figure 13-15.) 
The figure shows R and P to be 
distinct points. We shall prove 
that they are the same point. 

If Pi=R, then OR is the 
hypotenuse of right triangle 
AOPR and OR > OP, But R 
and P are both on C. Therefore 
OR = OP. Since this is a con- 
tradiction, it follows that P = R. 
Since I ±OP at P t it follows 
that I ± OR at R, and the proof 
is complete. 

Figure 13-15 

Our next theorem is the converse of Theorem 13.3. 

THEOREM 13.4 Given a circle and a line in the same plane, if 
the line is perpendicular to a radius at its outer end, then the line 
is a tangent to the circle. 

Proof: Assigned as an exercise. 

THEOREM 13.5 A diameter of a circle bisects a chord of the 
circle other than a diameter if and only if it is perpendicular to the 
chord. 

Proof: Let a circle C with radius r, center O, diameter AJ3, and chord 
RQ be given. Wc must prove two things. 

Ll£AB±KQ t then A75 bisects ftg. 
2. If AB bisects RQ, then A~B ± RQ. 

Proof of 1 : Choose an ^-coordinate system in the plane of the given 
circlc_C such that the origin is at the center 0,A = { - r, 0), B = (n 0), 
and RQ is perpendicular to KB at P(a, 0), where < a < r as shown 
in Figure 13-16, Then AB is a diameter of the circle. "Why? 



13.3 Tangent Lirm 561 




Figure 13-16 



Suppose /? and {) are named so that R is in the first quadrant. Then 
(see the proof of part 3 of Theorem 13,2) R = (a, y/& — a 2 ) 
Q = (a, — ^/r* — a 1 }. Therefore 



and 



FR = JVf«-fl 2 -0| = V* " o* 
Pp = p - (- Vf2 - a*)j = v^^ 1 ^- 



Therefore Fit = FQ and AB bisects ff^ at P. This proves that if a di- 
ameter is perpendicular to a chord, then it bisects the chord. 

Proof of 2; Choose an ary-coordinate system in the plane of the given 
circle C such that the origin O is the center of the circle, A = ( — r, 0), 
B = (r, 0), and the chord EQ (not a diameter) intersects AB at P{a, 0) 
with < a < r, (See Figure 13-17.) 




Figure 13-17 



562 Circles and Spheres Chapter 13 

Suppose that P is the midpoint of RQ. Then If R = (x^ ^) and 
Q — (** Sfe)* *■ nave 

Pfl = Pp, 
(*! - a)> + kl! = (* 2 - a ) 2 + y 2 2 , (Why?) 
*i 2 - 2*ia + 0* + y,a = Xi 2 - 2x& + a 2 + y#. 

But xi 2 + tji 2 = f 2 - *a a + yz 2 - (Why?) 

Then — 2x\a a — 2^ 2 a> *i = **. RQ is a vertical line, and RQ I 
AB, This proves that if a diameter bisects a chord that is not a diameter, 
then the diameter and the chord are perpendicular. This completes 
the proof of Theorem 13.5, 

THEOREM 13,6 In the plane of a circle, the perpendicular bi- 
sector of a chord contains the center of the circle. 

Proof: In the proof of part 1 of Theorem 13.5, AB is the perpendic- 
ular bisector of chord BQ in the plane of the given circle by the defi- 
nition of the perpendicular bisector of a segment in a plane. Since An 
contains O, the center of the given circle. Theorem 13.6 is proved* 

TIJEOREM 13.7 Let a circle C and a line I in the plane of the 
circle be given. If I intersects the interior of C, then I intersects C 
in exactly two distinct points. 

Proof-. Theorem 13.7 follows from part 3 of Theorem 13,2. The de- 
tails of the proof are assigned as an exercise. 

THEOREM 13,8 Chords of congruent circles are congruent if 
and only if they are equidistant from the centers of the circles. 

Proof: Let C" and C be the given congruent circles with centers P 
and F and radii f and f, respectively. Then r = f. Why? Let AB and 
A'B' be chords of the given circles C and C, respectively. Suppose, 
first that the distance from Ali to the center of circle C is zero, that is. 
AB is a diameter of C, and suppose that AU ^= A'B'. Then 

AB = A'B' = 2r = If 

and ~ATE is a diameter of circle C Therefore AB and A'B' arc equi- 
distant from P andF (th e dista nces being in this case). Conversely., if 
the distances of AB and A'B' from P and F are 0, then AB and /PF 
are diameters of C and C. It follows that AB = 2r = 2/ = A'B 1 ; 
hence AB === A'B', 



13.3 Tangent Lines 

Now suppose that AB is not a diameter of C. If AB s A'B\ it fol- 
lows that A'B' is not a diameter of C, Whv? Let F be the foot of the 

perpendicular from P to AB and let F be the foot of the perpendicular 
from F to A'B' as shown in Figure 13-18. Then by Theorem 13 .5, 
A** = l 2 AB and A'F = JA'B'. 



5*3 





Figure 1 3- IS 



By hypothesis, 
"Therefore 
We have 



AB = A'B'. 
AF = A'F. 
AP = A'F. 



(Why?) 



Since AAPF and AA'FF are right triangles, it follows that AAPF 
ss AA'FF by the Hypotenuse-Leg Theorem. Therefore PF = FF 
and hence AB and A'fr are equidistant from F and F, 

Conversely, if KB and A'B' are equidistant from P and F, that is, 
if 

Pf?=FF#0 ( 

ft follows that AAPFss AA'FF by die Hypotenuse-Leg Theorem, 

(Show that AAPF & AA'FF if PF = FF.) Therefore AF = A'F. 
But by Theorem 13.5, 



AF = 



and A'F = ^A'F. 



AB = A'B' and ^EssA'/?'. 
This completes the proof of Theorem 13.8, 

It should be noted that the congruent circles of Theorem 13.8 could 
be the same circle, in which case the theorem still holds. 



564 arete* and Spheres 



Chapter 13 



Figure 13-19 shows two different examples of two circles tangent 
to the same line at the same point In Figure 13-19& the centers A and 
A' of the two circles are on the same side of the tangent line I and the 
circles are said to be internally Umgent, In Figure 1 3- 1 9b the centers 
B and B' of the two circles are on opposite sides of the tangent line n 
and the circles are said to be externally tangent. 




[-Iml-.-ujIIj, Rutgeal - 

m 



Figure 13-19 

Our formal definition follows. 



Definition 13.8 l\vo circles are tangent if and only if they 
are coplanar and tangent to the same line at the same point 
If the centers of the tangent circles are on the same side of the 
tangent line, the circles arc said to be internally tangent. If 
their centers arc on opposite sides of the tangent line, the 
circles are said to be externally tangent. 



EXERCISES 13.3 



Exercises 1-10 refer to the circle C •= {(*, tf) : x 3 ■§■ tf = 64}. In each ex- 
ercise, the coordinates of a point arc given. Tell whether the point is en the 
circle, in the interior of the circle, or in the exterior of the circle. 



h (0, -8) 


4. (-4,7) 


7. (0, 8) 


10. (6. -2\/7) 


2, (3, 5) 


5. (-4v^,4) 


8. (8, -1) 




3. (-7,3) 


G, (4V3, -4) 


9. (-6, -6) 





11. Find the endpoints of two distinct diameters of the circle 

C = {(*» y) : ** + y* = ea } . 

12, Use set-builder notation to express the set of points in the exterior £ 
of the circle C = {(*, tj) : x-' + tf = 64}. 



13,3 Tangent Un*s 565 

13. Use set -builder notation to express the set of points in the interior I of 
the circle C ■ {(x, y) : x 2 + y 2 = 64}. 

14. Prove that the center of a circle is in the interior of the circle. 

Exercises 15-20 refer to the circle C" with center tit (0, 0) and passing through 
the point ?(- 5, 12). 

15. Find the radius of the circle. 

16. Use set-builder notation to express the set of points on the circle. 
17* Find eight distinct points that are on the circle 

18. Find two distinct points that are in the exterior of the circle. 

19. Find two distinct points that are in the interior of the circle. 

20. If / is the tangent line to C at P(— 5, 12). find an equation for L (Hint; 
I — UP at P, Find the slope of I and use the Point- Slope Form of an 
equation,) 

21. Ut the circle C = {(*, y) : x 2 -f y* = 36} and the line 
I = {(x, y) : x = 3} be given, 

(a) Does I intersect C? 

(b) If / intersects C, is I a tangent line or a secant line? 

(c) If I intersects C, find the coordinates of the poiutjs) of intersection. 

22. Let the circle C = {(x, y) : x* + y 2 = 25} and the lines 

Ei = {(*.y) : x = -5}, h= {(*,$f) : y = 2*}, h = {fry) : * = 7) 
he given. 

(a) Which of the lines intersect the circle? 

(b) Which line is tangent to the circle? What are the coordinates of the 
point of Langency? 

(c) Which line is a secant? What are the coordinates of the points of 
intersection of the secant line and the circle? 

23. Prove Theorem 13.4. 

24. Prove Theorem 13.7, 

25. Copy and complete: A tangent to a circle is JT] to the radius drawn to 
the point of contact. 

26. Copy and complete: If a diameter is perpendicular to a chord, then it 
[?] the chord. 

27. Copy and complete: If a diameter bisects ft chord other than a diameter, 
then it Is |3. 

28. Copy and complete: In the plane of a circle, the perpendicular bisector 
of a chord contains the [T]. 

29. In a circle with r ad his 13 in., how long is a chord 5 in. from the center 
of the circle? 

30. In a circle with diameter 12 cm., how long is a chord 4 cm, from the 
center of the circle? 

3L Find the radius of a circle if a chord 8 in. long is 3 in. from the center of 
the circle. 



566 Circles and Spheres 



Chapter 13 



32, How far from the center of a circle with a radius equal to 25 is a chord 
whose length is 30? 

33. In the figure. FQ is parallel to I which is tangent to the circle at F. The 
center of the circle is R and FQ bisects RF at Jtf. If FQ = IS, find BP, 
[Hint: Let RF = 2x. Then RP = 2x and RM = x.) 



w 


Jl 



34. If AH is a diameter of a circle and if lines t t and i-* arc tangent to the 
circle at A and B t respectively, prove h \\ £&. 

35. Prove that the line containing the centers of two tangent circles con- 
tains the point of tangeucy. (See Figure 13-19.) 

36. The figure below at left shows two concentric circles, AB is a chord of 
die larger circle and is tangent to the smaller circle at AC Prove that M 
is the midpoint of A~B t 





37. In the figure above at right, T is a point in the exterior of the circle with 
center P. T\vo distinct tangents are drawn from T to the circle with 

points of contact A and B. Prove that TF bisects LATB and that 
AT = BT. 



3S. In Exercise 37. if the radius of the circle is 9 and FT = 15, find AT. 
39. challenge problem. In Exercise 37 if the radius of the circle is 9 and 
FT = 15, find AH. 



13.4 Tangent Plana* 567 

40. CHALLENGE PROBLEM. Show thilt the lillC I = {(x. If) I 3x 4- 4(/ = 25} 

i8 tangent to Uic circle C = {(*, tf) : ** + y 2 = 25) and find the co- 
ordinates of the point of tangency. 

41. challenge problem. Prove that no circle contains three col linear 
points. {Hint: Use Theorem 13.6.) 



13.4 TANGENT PLANES 

In Section 13.3 we studied relations between lines and circles in a 
plane. In this section we study relations between planes and spheres in 
space. There is a close analogy between the definitions and between 
the theorems of the two sections. 

Onr first theorem of this section is analogous to Theorem 13.1, In 
Chapter 12 you learned how to find the distance between two points in 
space by introducing a three-dimensional coordinate system (called an 
$j/£-coordinatc system). We shall use this Distance Formula to develop 
an equation for a sphere in an xy£-coordtaate system. 

Let O be a point and r a positive numl ier. Let S be the sphere with 
center O and radius r. Suppose an ^-coordinate system has been set 
up with O as the origin. (See Figure 13-20.) Then Pfc tj, %) is a point of 
S if and only if 

OF = s/{x - 0)2 + (y - 0)8 + (Z - 0)-' = r s 

V* 2 + tf 3 + « a = r, 
* a + y 2 4- z 2 = r 2 . 



P(x,y,»ji 




Figure 13-30 



We have proved the following theorem. 



568 Circles and Spheres 



Chapter 13 



THEOREM 13.9 Let be a point, r a positive number, and S the 
sphere with center O and radius r. Given an jcy.s-coordmate system 
with origin O, S = {(ar, y, z) ; x 2 + y 2 + s 2 = r 2 }. 

Definition 13M (See Figure 13-21.) Let a sphere with 
center and radius r be given. The interior of the sphere is 
the set of all points P in space such that OF << r. The exterior 
of the sphere is the set of all points P in space such that 
OP>r. 




e 



J:.-,'Tk,r 



*\. Exterior 



<> 



-^ 



Figure 13-21 

In view of Definition 13-9 and Theorem 13.9. if S is a sphere with 
radius r and center at the origin of an 3fys-coordinate svstem, then 

S = {(*, if, z) : X 2 + tf + *2 = r 2 } 

Z= {(*,y f *):a* + if + ^<ia} 
and 

where / is the interior of the sphere and £ is its exterior. 

Before reading Definition 13.10, try to form your own definition 
of a plane tangent to a sphere. See Figure 13-22. 



Figure 13-23 




13.4 Tangent Plants 569 



Definition 13.10 If a plane intersects a sphere in exactly 
one point, the plane is called a tangent plane to the sphere. 
The point is called the point of tangent- \\ or the point of 
contact, and we say that the plane and the sphere are tangent 
at this point. 



You have seen that there are three possibilities with regard to the 
intersection of a line and a circle in the same plane. Figure 13-23 
suggests three possibilities with regard to the intersection of a plane 
and a sphere. For each part of Figure 13-23. S is a sphere with center O 
and P is the foot of the perpendicular from O to plane a. Figure 1 3-23a 
suggests that if P is in the exterior of the sphere, then all of plane a is in 
die exterior of the sphere and S n n = . Figure 13-23b suggests that 
if jP is on the sphere, then a is tangent to the sphere at P and Sfla = 
{P}. Figure l3-23c suggests that if P is in the interior of the sphere, 
then the intersection of the sphere and plane ot is a circle C with center 
P, that is, S n a = C. 



a 





(a) 



(b) 



(O 



Figure 13-33 

The following theorem concerning the intersection of a plane and 
a sphere is analogous to Theorem 13.2 concerning the intersection of a 
line and a circle in a plane. It can be proved using a three-din ie us it mal 
coordinate system in much the same way that Theorem 13.2 was 
proved using a two-dimensional coordinate system. You will be asked 
to write a proof of Theorem 13.10 in the Exercises. 

THEOREM 13.10 Given a sphere S with center O and a plane a 
which docs not contain O, let P be die foot of the perpendicular 
from O to a. 
L Every point of a is in the exterior of S if and only if P is in the 

exterior of S. 
2* a is tangent to S if and only if P is on $♦ 
3. a intersects S in a circle with center P if and only if P is in the 

interior of S, 



570 Circles and Spheral Chapter 13 

1HEOREM 13. J J Let a sphere S with center O and radius r and 
a plane a be given. If the intersection of S with a contains the center 
O of the sphere, then the intersection is a circle whose center and 
radius are the same as those of the sphere. 

Proof: Let a sphere S with center O and radius r be given as shown in 
Figure 13-24. Let a be a plane that contains O and intersect*. S. Since S 
is the set of all points of space whose distance from O is r, the inter- 
section of S and a is the set of all points P of a such that OP = r. By 
definition, this set is the circle in plane a with center O and radius r. 
Thus the circle has the same center and radius as the sphere, and the 
proof is complete* 




Figure 13-24 

Definition 13.11 A circle that is the intersection of a sphere 
with a plane through the center of the sphere is called a great 

circle of tile sphere. 



The next two theorems can be easily proved using Theorem 13, 1 1. 

THEOREM 13.12 The perpendicular from the center of a sphere 
to a chord of the sphere bisects the chord. 

Proof: The endpoints of the given chord and the center O of the given 
sphere determine a plane O. The intersection of plane a with the sphere 
is a great circle: with center O and having the same chord as the given 
chord- It follows from Theorem 13.5 that the perpendicula: Irom ( Mo 
the given chord bisects the chord. 

THEOREM 13.13 The segment joining the center of a sphere to 

the midpoint of a chord of the sphere is perpendicular to the chord. 

Proof: Assigned as an exercise. 



13.4 Tangent Planes 571 

Our last theorem of this section is analogous to Theorems 13.3 

and 13.4. 

THEOREM 13. 14 A plane is tangent to a sphere if and only if it 
is perpendicular to a radius of the sphere at its outer end. 

Proof: Let ft sphere S with center O and radius OF be given. Let « 

tie the given plane. There are two parts to the proof. 

1. If UP A a at P, then a is tangent to S at P. 

2. If a Is tangent to 5 at P y then OP _ a. 

Proof of h We are given that OP J- a al P as shown in Figure 13-25. 
Let R be any point of a different from P; then OP X M (Why?) and 
AOPR is a right triangle with the right angle at P. Therefore 
OP < OR. Why? Therefore ii is a point in the exterior of S. Why? It 
follows that P is the only point of a that belongs to boil a and S; hence 
0! is tangent to S at P. 




Figure 13-23 



Proof of 2: We are given that a is tangent to S at P. We shall use an 
indirect proof to show that OP ± a, Suppose, contrary to what we 
want to prove, that OP is not perpendicular to a. Let Q be the foot of 
the unique perpendicular from O to a. Then OQ < OF. Why? There- 
fore Q is in the interior of S, Why? It follows from part 3 of Theorem 
13.10 that a intersects S in a circle. But this contradicts the hypothesis 
that a is tangent to S; that is, the intersection of a and S is exactly one 
point. Therefore our supposition that UP is not perpendicular to a is 
incorrect and we conclude that OP _L a. This completes the proof of 
Theorem 13.14, 



572 Circles and Spheres Chapter 13 

EXERCISES 13.4 

L I*et a sphere with radius 10 in, lie given, A plane 6 in. from the center 
of the sphere intersects the sphere in a circle. Find the radius of the 
circle. 

2. Given a sphere S with center O, let a and B be two planes equidistant 
from O and such that a intersects S in a circle C and /? intersects S in a 
circle C. Prove that C is congruent to C. {Hint; Let F be the foot of the 
perpendicular from O to o and let F be the foot of the perpendicular 
from O to jS. Let G be a point on C and C" be a point on C. Prove that 
&Omm &OFC.) 

3. In Exercise 2, is it necessary for « and /? to be parallel planes? 

4 Prove that if the circles of Intersection of two planes with a sphere are 
congruent, then the planes are equidistant from the center of the 
sphere. 

5, Let a sphere with radius 12 be given, A segment from the center of the 
sphere to a chord and perpendicular to the chord has length 8, Find the 

length of the chord, 

§. A sphere with center C is tangent to plane a at P. AS and CD are lines 

in plane a which contain P. In what way is CP related to AJS? To CO? 
Draw a figure which illustrates the given information. 

7. Prove Theorem 13.13. {Hint: Give a proof similar to that of Theorem 
13,12.) 

8. Given a sphere S with center P as shown in the figure. HAS and CD are 
chords of S which are equidistant from F, prove that 

AB 5s W and LABP m Z CDP. 

{Hint: Use Theorem 13.8.) 




Given that AC and BD are perpendicular diameters of a sphere, prove 
that ABCD is a square. 



13.4 Tangent Planet 573 



10. Given that AC and BD are distinct diameters of a sphere, prove that 
ABCD is a rectangle. 

11. Let a sphere S with center P as shown in the figure be given. C is a great 
circle of S. R is a point on C and T is a point on S, but T is not on C. If 
ml HPT = 60, prove that ARPT is equilateral. 




12. Let a sphere H and a plane a tangent to S at point A be given. Let 
plane be any plane other than a which contains A. (See the figure,) 




(a) Prove that plane fi intersects sphere S in a circle C. 

(b) Prove that piano /? intersects plane a m a line /. 

(c) Prove that t is tangent to C at A. 

(Hint: Suppose I intersects C in a second point Q, Then Q is on S 
(Why?), and hence a intersects S in a second point Q. Contradiction?) 



574 Circles and Spheres Chapter 13 

■ Exercises 13-22 refer to the sphere 

S= [&$$%& + & +&mM}< 

In each exercise, given the coordinates of a point, tell whether the point is 
on the sphere, in the interior of the sphere, or in the exterior of the sphere. 

13. (0,0,8) 18* (-6, -4,2V3) 

14. (0, -8,0) 19. (4,-4, -8) 

15. (4 3, 5) 20. (5, 0, -6) 

16. (7,2,3) 21. (_2,2 X /B,0) 

17. (-4,5,6) 22. (4.3V$ -2) 

23. Find the endpoints of three distinct diameters of the sphere 

$= {{x f if,z):x* + f + J = lQa). 

■ Exercises 24-28 refer to the sphere with center at (0, 0. 0) and containing 
the point (4 t 4* 2). 

24. Find the radius of the sphere. 

25. Use set-builder notation to express the set of points an the sphere, 

26. Find the coordinates of two distinct points that are on the sphere. 

27. Find the coordinates of two distinct points that are in the exterior of 
the sphere. 

2& Fiod the coordinates of two distinct points that are in the interior of 
the sphere. 

29. challenge PROBLEM. Theorem 13.14 could be called a restatement 
of part 2 of Theorem 13.10* Complete the proof of the following restate- 
ment of part 3 of Theorem 13.10. 

The intersection of a plane and a sphere is a circle whose center 
is the foot of the perpendicular from the center of the sphere to the 
plane if and only if the foot of the perpendicular is in the interior of the 

sphere. 




30. 



13.4 Tangent Plane* 575 

Proof; 1 ,et a sphere S with center O and radius r and a plane a be 
given. Let P he the foot of the perpendicular from O to a as shown in 
the figure. There are two things to be proved, 

(a) If OP < r, then or H S is a circle G with center P. 
(1>) If a n S is a circle with center F, then OF < r. 

Let X be any point of the intersection of a and S. To complete the proof 
of (a) you need to show that FX is a constant for all points X in the in- 
tersection of a and 5. To complete the proof of (b) you need lo show thai 
if FX is a constant for all points X in the intersection of ct and 8, then 
OP<r. 

CHALLENCF problem. Complete the proof of the following restate- 
ment of part 1 of Theorem 13.10. 

The intersection of a plane and a sphere is the empty set if and only 
if the foot of the perpendicular from the center of the sphere to die 
plane is in the exterior of the sphere 

Proof: Lei a sphere S with center O and radius r and a plane « be 

given. Let P be the foot of the perpendicular from O to <* as shown in 
the figure- 




There are two things to be proved. 

(a) If OP > r t then all points of « are in the exterior of the sphere. 

(b) If all points of « are in the exterior of the sphere, then OP > r, 

31. challenge problem. See the proof of Theorem 13,2. Prove Theorem 
13.10 in a similar way using an xip-coordinate system. 



576 Circle and Spheres 



Chapter 13 



13,5 CIRCULAR ARCS. ARC MEASURE 



Thus far in this chapter we have treated circles and spheres in a 
similar maimer. In the remainder of this chapter, we limit ourselves to 
the consideration of topics relating to circles only. The reason for this 
is that the treatment of the corresponding topics for spheres is too com- 
plicated to consider in a first course in geometry. We begin with some 
definitions. 



Definition 13.12 An angle which Is coplanar with a circle 
and has its vertex at the center of the circle is called a central 
angle. 




Figure 13-S6 



In Figure 13-26, P is the center of Ihe given circle and AT) is a di- 
ameter. LAFB is a central angle. Name three more central angles 
shown in trie figure. 



Definition 13.13 (See Figure 13-27.) If A and B are distinct 
points on a circle with center P and if A and B are not the end- 
points of a diameter of the circle, then the union of A, B, and 
all points of the circle in the interior of LAPB is called a 
minor arc of the circle. The union of A, B, and all points of the 
circle in the exterior of LAPB is called a major arc oi the 
circle. If A and B are the endpoiuts of a diameter of the circle, 
then the union of A, B, and all points of the circle in one of the 

two halfplanes, with edge AB t lying in the plane of the circle 
is called a semicircle. 



13,5 Circular Arcs, Arc Measure 577 




^i ic."n 



Figure 10-27 

From Definition 13.13, an arc of a circle is either a minor arc, a 
major arc, or a semicircle. The points A and B in Definition 13.13 are 
called the endpoints of the arc. 

Notation. We may denote an arc with endpoints A and B by the sym- 
bol AjB, which is read "arc AB." However, it should be noted that the 

symbol AB is ambiguous unless the word "minor" or the word "major" 
is used in connection with the symbol. For example, in Figure 13-27, 

we may speak of the minor AjB or the major Aft* Also, semicircle AB 
is ambiguous since there are two semicircles with endpoints A and B. 
One way to avoid confusion as to which arc is meant is by choosing an 
interior point of the arc in question (tliat is, a point of the arc other 
than its endpoints) and using this third point in naming the arc. Thus, 

in Figure 13-28, w© may speak of minor AXB or simply AXB. Similarly, 
we may speak of major A YB or simply A YB, semicircle CXB or semi- 
circle CYB. 




Figure 13-2S 



=78 Circles and Sphtret C-iap:^ 13 

It is clear that for each pair of distinct points A, B on a circle there 
are two arcs which have these points as endpoints. If AB is a minor are, 
we sometimes say that major AB is tho corresponding major are, or 
that major AB corresponds to minor AB, or that minor AS corresponds 
to major AB, 

In Figure 13-28 name the major arc that corresponds to minor BX. 
Name the minor arc that corresponds to major ABY. 

When we speak of the measure of a segment, we mean the nnm- 
bcr that is the length of the segment. However, this is not true of arcs. 
That is, when we speak of the measure of an are, we do not mean the 
"length" of the arc, since length has not been defined for anything 

except segments. If AXB is a minor arc of a circle with center F, we 

say that / APB is the associated central angle with respect to AXB. 
The measure of a minor arc of a circle is related to the degree measure 
of its associated central angle. We make the following definitions. 



Definition 13, 14 II AXB is any arc of a circle with center P t 

men its degree measure (denoted by mAXB) is given as 
follows: 

1. If AXB is a minor arc, then mAXB is the measure of the 
associated central angle; that Is, 

mAXB = mZ APR 

2. If A XB is a semicircle then 

tnAXB = ISO. 

3. If AXB is a major arc and AY& is the corresponding minor 
arcs then 

mAXB = 360 - niAYB, 



Figure 13-29 shows a circle with center i J . If 
mlAPB = 50, 



then 
and 



mAXB = 50 

mXyS = 360 - 50 - 310. 



13.5 Circular Arci, Arc Measure 



57*3 




Figure 13-19 

If BC is a diameter of the circle, what is mCXB in the figure? What is 
mCYB? 

Hereafter we shall call mAXB simply the measure of arc AXB with 
the understanding that wc mean the degree measure of the arc. Is the 
measure of a minor arc always less than 1 80? Why? Ts the measure of 
a major arc always greater than 180? Why? Can the measure of an arc 
be zero? Why? 

Note that the measure of an arc does not depend on the size of the 
circle which contains the arc. Figure 13-30 shows 'Concentric circles 
with center P. with 

mCYD = mAXB = mLAVB = mlCPD = 35. 




Figure 13-30 

Given three points A, B, C such that B is between A and C, we 
know that 

AB 4- BC = AC. 
Suppose we are given an arc ABC (that is, B is a point, but not an end- 
point, of AC). It seems reasonable that 

inAB + JtiBC = rnAFC. 

We state this as our next theorem. 



5 BO Circtas and Spheres Chapter 13 

THEOREM 13.15 {Arc Meamre Addition Theorem) If A, B, C 

are distinct points on a circle, then 

mXlid = mAB + mBC 

Proof; Followi ng is a plan for a p roof considering seven possi ble cases 
as shown in Figure 13-3 1 . In each case, V is the center of the circle and 
the assertion of the theorem follows from the listed equations. This 
plans a complete proof since there are no other cases. 





c A\ 




U-.-C J 



Ciae2 



Cosu U 









Cut? 



Cast I Quti 

Hgur* 1331 

Case 1, ABC is a minor arc. Then AB and BC are minor arcs, 

mABC = mlAVC 
mAB = ml AVB 
mlRm ml BVC 
m I AVC = m I AVB + ml BVC. 

Case 2. ABC is a semicircle. Then A B and BC are minor arcs, 
mABC = 180 
mAB = mlAVB 
mBC- ml BVC 
180 = m I AVB -+- m L BVC. 

3. ABC is a major arc, and AB and BC are minor arcs. Then 
mABC = 360 - mlAVC 
mAB = m I AVB 
mBC = ml BVC 
360 = ml AVB + ml BVC + mlCVA. 



13.5 Circular Arcs, Arc Manure a 81 

Case 4. ABC is a major arc, AB is a minor arc, and BC is a major arc. 
Then 

mABC = 360- mZ.AVC 

mAB = mLAVB 

mBC = 260- mLBVC 

mLBVC = mLBVA + mLAVC. 

Goa& 5. AJ3C is a major arc, AB is a minor arc, and BC is a semicircle. 
Then 

mABC = 360 - mLAVC 

mAB = mLAVB 

m£C= 180 

180 = mZAVB + mLAVC 

Case 6. ABC is a major arc, AB is a semicircle, and BC is a minor arc. 
Then 

mABC = 360- mLAVC 

mAB = 180 

mBC = mLBVC 

180 = mLAVC + mLBVC. 

Case 7. ABC is a major arc. AB is a major arc, and BC is a minor arc. 
Then 

mABC = 360- mLAVC 

mAB = 360 -mLAVB 

mBC = mLBVC 

mLAVB s mLAVC + mLCVB. 

Note in the situation of Theorem 13, 15 that if D is a point of the 
circle not on ABC, then 

mCD/1 = 360 - mABC 
and 

mAB + mBC + mGDA = mA5c + mCDA = 360. 
In other words, if A. B, C are three distinct points on a circle that par- 
titions the circle into arcs, AB, BC, CA, intersecting only at their end- 
points, then 

mAB + mBC + inCA = 



582 Circles and Spheres Chapter 13 

This idea may be extended to any numl>er of points on a circle, as in 
the following corollary. 

COROLLAR Y I.'i, 15,1 U A i, A 2t . , . , A E are n distinct points on 
a circle that partition the circle into n arcs 

AjA 2 + AvAz* • . . * A„_iA B> A^Ai 

that intersect only at their endpoints, then 

tnAiA 2 + mAvAs + ■ • * + mA n _iA n + ntA Ji A 1 = 360. 

Proof: Following is Lhe proof for n = 5, The proofs for other values 
of n arc similar. By repeated application of Theorem 13.15, we get the 
following: 

mA L A 2 + mAzA s 4- mAfiA* 4- mA^Aa + mA^A L = 

mAiA'jAy -f- mAaA4 + 771A4A5 + mA$A l = 

mAiA^A* 4- rnA+As - mA$A\ = 

mAiA 2 A 5 -r mAgAi = 360. 

EXERCISES 13,5 

■ Exercises 1-7 are on the proof of Theorem 13,15. In each exercise, show 

how to derive the assertion oJ the theorem from the listed equations for the 
given case. 

1. Case 1 5. Case 5 

2. Case 2 6. Case 6 

3. Case 3 7. Case 7 

4. Case 4 

8. In the figure, mABC = 240 and mBA'C = 100. Find mAZB and ntAYC. 

A ^ -^ Z 




13.5 Circular Arcs, Arc Measure 583 
Exercises 9-20 refer to Figure 13-32. In the figure, P is the center of the 

circle; A, B, C, are points on the circle; A-P-B-, mBC — 50; and mBD =s 

110. 




no 

Figure I&32 

9, Name five minor arcs determined by points labeled in the figure. 

10. Kind tile measure of each of the minor arcs named in Exercise 9. 

11. Name the five major arcs corresponding to the five minor arcs named in 
Exercise 9, 

12. Find the measure of each of the major ares named in Exercise 11. 

13. Name two semicircles. 

14. Which theorem justifies the conclusion thai mCBD = 160? 

15. Find m LBFC. 

16. Find m L LTD. 

17. Copy Figure 13-32 and draw A£, BC t and FC. Find mlBAC, How 

does m L BAC compare with mBC? 
ia Show that mlACB = 90. 

19. Draw A~D, BD, and PD on your copy of Figure 13-32. Show that 

ml BAD = [mBlX 

20. Find m£ADB. 

Exercises 21-23 refer to Figure 13-33. In the figure, Pis the center of the 
circle; A, B, C are points on the circle; and A-P-B. 

21. Copy Figure L3-33 and draw FC. 
Prove thiil ml BAC = \mBC, 

22. Prove thai in L ABC = fynAC. 

23. Prove that A ABC is a right 
triangle. 



Figui* 1*33 




584 Circles and Spheres 

24. CHAIJ.ENGE PROBLEM. Ill the KgUTe, 

Pis the center of the circle; A t B, 
C, D arc points on the circle; 
and A-F-B. Prove that 

m£CAD = %mCBD. 




13.6 INTERCEPTED ARCS, INSCRIBED ANGLES, ANGLE MEASURE 

In the discussions that follow we shall he concerned with angles in- 
scribed in an arc of a circle and about arcs of a circle that are inter- 

cepted by certain angles. In Figure 13-34a, £ABC is inscribed in ABC 

and t ABC intercepts AXC In Figure l&34b, £PQR is inscribed in 

PQR and Z FQR intercepts FYR. 




Figure 13-34 

None of the angles shown in Figure 13-35 is an inscribed angle, but 
each angle intercepts one or more arcs of a circle. In Figure 13-35a, 

AAFB intercepts AXB. In Figure 13-35h, Z R ST intercepts RYS, In 

Figure 13.35c £AVB intercepts AKR and also CMD. In Figure 

13-35d, Z GEF intercepts GHF, In Figure 13-35C, Z A VB intercepts 

ARB and also CSB, In Figure 13-35f t £AVB intercepts Ai5i and also 

AGB. 



13.6 Intercepted Arcs, Inscribed Angles, Angle Measure 585 




Figure 13-35 



On the other hand. L LKP 
shown in Figure 13-36 does not 
intercept an are of the circle. 



Figure 1U-H0 




586 Circlet and Spheres 



Chapter 13 



Our formal definitions of an Inscribed angle and of an intercepted 
arc follow. 



Definition 13.15 An angle is said to l>e inscribed in an arc 
of a circle and is called an inscribed angle if and only if both 
of the following conditions are satisfied ; 

1 , Each side of the angle contains an endpoint of the arc. 

2. The vertex of the angle is a point, but not an endpoint, of 
the arc. 



Explain why each of the angles shown in Figure 13-35 fails to be an 
inscribed angle. Explain why each of the angles shown in Figure 13-34 
is an inscribed angle. Draw a picture of an angle inscribed in a 
semicircle. 



Definition 1X16 An angle is said to intercept an arc of a 
circle and the arc is called an intercepted arc of the angle if 
and only if all three of the following conditions are satisfied: 

1. Hie end points of the arc lie on the angle. 

2. Each side of the angle contains at least one endpoint of the 



arc. 



3. Each point of the arc, except its endpoints, lies in the in- 
terior of the angle. 



You should check to see that each of the figures shown in Figure 
13-35 satisfies all three of the conditions stated in Definition 13.16. 
Which of the tlircr- conditions stated in Definition 13.16 are not satis- 
fied by the figure shown in Figure 13-36? 

Figure 13-37 shows two angles inscribed in the same arc. (Of 
course, the angles intercept the same 
arc) Name the arc in which both 
angles, LABD and £ ACD, are in- 
scribed. Name the arc that both of 
these angles intercept, it appears 
that Z ABD and LACD in the figure 
are congruent. (Measure each of 
them with your protractor,) That 
they actually are congruent is a cor- 
ollary of our next theorem. 

Figure 13-37 




13.S Intercepted Arcs, Inscribed Angles, Angle Measure 587 

THEOREM 13.16 The measure of an inscribed angle is one-half 
the measure of its intercepted are. 

Proof: Let a circle with center V be given and let Z ABC be inscribed 
in ABC. Then the intercepted arc is AC* We must prove mLABC = 
JmAC There are three possible cases as suggested in Figure 13-38. 




Casel. VisonZA^C. 

Case 2. V is in the interior of Z ABC, 

Case 3. V is in the exterior of /ABC. 



Figure 13-38 



Proof of Case I: Since V is on Z ABC, then either AB or BC is a di- 
ameter of the given circle. Suppose BC is a diameter as in Figure 
I3-3&L Draw VA Then AAVB is isosceles (Why?) and mLA = 
m it ABC /.AVC is an exterior angle of AAVB, so by Theorem 7-30, 

m/AVC = mlA + mLABC 

Since m£A = m£ABC r 

n-UAVC = mLABC + mZA3C, 



or 

But 

Therefore 



2?n£ABC = mlAVC, 

m£AVC = mAC (Why?) 

2mZABC= mAC 

mLABC = ^niAC. 
This completes the proof of Case 1 . 

Proof of Case 2: V is in the interior of /.ABC as shown in Figure 

13-38b, so B Vis between BA and BC. Let B' be the point where opp VB 
intersects the circle; then BW is a diameter and B' is an interior point 



5SB Circles irwj Spheres Chapter 13 

of the intercepted arc AC. It follows from Case I that 

m I ABB' = $mAB' and ml B'BC = ^mBC 
Adding, we get 

ml ABB' + m/B'BC = \mAB> + ^nB~C, 

or 

ml ABB' + mlWBC = J(w*dB' + mB'C). 
But 

ml ABB' + ml B'BC = ml ABC (Why?) 
and 

mAB' + miPc = wtAC : Why?) 
Therefore 

mlABC= fytiAC 

and the proof of Case 2 is complete. The proof of Case 3 is assigned as 
an exercise. 

Recall how congruent segments and congruent angles are defined 
in terms of their measures. How would you define congruent arcs? 
Write what you think is a good definition of congruent arcs. Turn back 
to Figure 13*30. Does your definition allow the conclusion that arcs 

AXB and CYD of Figure 13-30 arc congruent? If it does, you should 
reword your definition so that it excludes this conclusion. Compare 
your definition with the following. 

Definition 2,127 Two arcs (not necessarily distinct) are 
congruent if and ony if they have the same measure and are 
arcs of congruent circles. 

Notation H arcs AB and CD are congruent, we write AB = CD. 

Note mat we cannot say that two arcs arc congruent unless we 
know that they have the same measure and we know that they are in 
the same circle or that they are in congruent circles. 

The following three corollaries are important consequences of The- 
orem 13.16, Their proofs are easy and are assigned as exercises. 

COROLLARY l&W.l Angles inscribed in the same are are 
congruent. 

COROLLARY 13.16.2 An angle inscrilied in a semicircle is a 
right angle. 



13.6 Intercepted Are*, Inscribed Angles, Angle Measure 539 

COROLLARY 13.16.3 Congruent angles inscribed in the same 
circle or in congruent circles intercept congruent arcs. 

Figure 13-39 shows two distinct purullel lines intersecting a circle. 
We call an arc whose endpoints arc on the lines, one endpoint on each 
of the lines and all of whose interior points are between the lines, an 

intercepted arc. Thus, in Figure 13-39, lines An and CD intercept arcs 

AXC and BYD. Are these two arcs congruent? They are congruent by 
our next theorem. 




Figure 13-39 

THEOREM 13.17 If two distinct parallel lines in the plane of a 
circle intersect that circle, they intercept congruent arcs. 

Proof: Let distinct parallel lines / and in intersect a given circle in 

arcs AXC and BYD. There are three possible cases as suggested in 
Figure 13-40. 




B B 



Figure 13-40 

Case L Both I and m are secant lines as shown in Figure 13-40a. 
Case 2. The line if is a tangent line and the other line m is a secant 
line as shown in Figure 13~40b. 
Case 3. Both I and m are tangent lines as shown in Figure 13-4Qc, 



590 Circles arui Spheres Chapter 13 

Proof of Case 1: Since / and m are secant lines, I intersects the circle 
in distinct points A and B, and m intersects the circle in distinct points 

C and D. (See Figure 13-40a.) We must prove that AXC m BYD. Draw 
B€. Then 



and 
But 

Therefore 
or 



mlBCD = 
ml ABC = $mAXC. 
mlABC = m L BCD. Why? 
\mAXC s JmBY/3 
mAXC = mBYD. 

AXC B BYD. Why? 

This completes the proof of Case 1 of Theorem 13,17, The proofs 
of Cases 2 and 3 are assigned as exercises, 

EXERCISES 13.0 

Exercises 1-17 refer to Figure 13-4L In the figure* P is the center of the 
circle: A, B, C, D arc points on the circle; AC is a diameter; mAD = 100; 
and mBC b 40. 




Kipiw 13-41 

1. Kamc four inscribed angles and name the arc each intercepts. 

2. Name the five minor arcs determined by points labeled in the figure. 

3. Name the five major arcs corresponding to the five minor arcs named 
in Exercise 2. 

4. Name L\vr> semicircles. 

5. Find the measure of each of the minor arcs named in Exercise 2, 
CJ. Find the measure of each of the major ares named in Exercise 3. 



13.6 Intercepted Arc*, Inicribed Anglei, Angle Measure 

7. Find the measure of each of the angles ZA, Z B, L C t and Z0. 

8. Find the measure of £CFD, 

9. Find the measure of LABC. 

10. Name an angle thai [S congruent to ZB, 

11. Name an angle that is congruent to LD, 

12. Prove that AD 1 CD. 

13. Find the measure of Z DCB. 

14. II AC intersects BD at K, name a pair of similar triangles, 

15. If AC intersects BD at E, prove that m / DEC = %{mCD + mAU). 

16. Prove that AADE - AW(7 

17. Prove that DE ■ £# = AE - EC, 



591 



18. With each chord of a circle which is not a diameter there are two asso- 
ciated arcs of the circle. One of the arcs is a minor arc and the other arc 
is a major arc. The endpoints of the chord are the end points of the arcs. 
Complete the proof of the following theorem. 

THEOREM 13. IS In the same circle, or in congruent circles, two 
chords that arc not diameters are congruent if and only if their asso- 
ciated minor arcs are congruent. 

Proof: Using the notation of the figure, we are given two congruent 
circles , C a nd C, with centers P and F, respectively. AB is a chord of 
C and A'W is a chord of C. There arc two things to prove. 




(a) If AB ~ A'B'. then A B tt A'B', 

(b) If AB s j?B', then M a A^F. 

{Hints In proving (a), show that AAPB s AAT7*' by the S.S.S. Pos- 
tulate. In proving (b), show that AAPJ? at AA'P'B' by the S.A.S. 
Postulate.'! 



592 Circles and Spheres 



Chapter 13 



19. Does Theorem 13.18 of Exercise 18 still hold if we replace "minor arcs" 
with "major arcs" in the statement of the theorem? 

20. Prove Case 3 of Theorem 13.16. (Refer to Figure 13-38c.) 

21. Prove Corollary 13.16.1. 

22. Prove Corollary 13.16.2. 

23. Prove Corollary 13.16.3. 

24. Figure 13-42 shows three cases of a chord AT of a circle with center P 

and a tangent VTto tlie same circle intersecting at the point of taugency 

V. Using the notation of the figure, the angle whose sides are rays VT 

and VA is sometimes called a tangent-chord angle. Complete the proof 
of the following theorem. 

THEORtlM 1^.19 The measure of a tangent-chord angle is one-half 
the measure of its intercepted arc. 

Proof: Using the notation of Figure 1 3-42, P is the center of the given 

circle, AXV is the intercepted arc, AT'is the given chord, and VTis the 
given tangent. There are three cases to consider. 




Figure 13-48 

Case L P is on VA as shown in Figure 13-42u. 

Case 2. P k in the exterior of I AVT as shown in Figure 13-42b. 

Case 3. P is in the interior of I A VT as shown in Figure 1 3-42c. 

Proof of Casn h (See Figure 13-42a,} AV is a diameter; hence 

mZ AVT = 90 (Why?) and mAXV = 180. Thus ml AVT = fynAXV. 

Proof of Case 2: (See Figure 13-42b.) Draw diamotcr VB. 

mlBVl = imHXV, Why? 

= limM +mAXV), Why? 
= {niffi + tyiAX$. 

Since mlBVT = mlBVA +mZ AVT, we have 

ml BVA + mlAVr = tynBA + intAXV 

fey the Substitution Property of Equality. Rut 



13.6 Intercepted Arc*, inscribed Angles, Angle Measure 593 

mlBVA = imBA. Why? 
Therefore, by the Addition Property of tonality, we get 

mlAVT=%mAXV, 
Complete the proof of Theorem 13,19 by proving Case 3, 

25. Prove Case 2 of Theorem 13.] 7. 

20. Prove Case 3 of Theorem 13.17. {Hint: In Figure 13-40c, prove that W 
is a diameter. It will then follow that HXD and BYD are semicircles 

and that BKD S BYD.) 

27. A quadrilateral is said to he inscribed In u circle and is called an in- 
scribed quadrilateral if all of its vertices are on the circle. Prove that the 
opposite angles of an inscribed quadrilateral are supplementary. 

28. If the diagonals of an inscribed quadrilateral are diameters, prove that 
the quadrilateral is a rectangle, (See Kxerci.se 27.) 

29. Prove that the midray of a central angle of a circle bisects the arc inter- 
cepted by the angle. 

30. The figure shows two secants intersecting in an exterior point V of a 

circle. The rays VA and VC are sometimes called secant-ray?; and the 
angle whose sides are these rays is called it secant-secant angle. Com- 
plete the proof of the following theorem. 

THEOREM 13.20 The measure of a secant-secant angle is one-half 
the difference of the measures of the intercepted arcs. 




Using the notation of the hgure t we must prove 

fliZV = JrimBYD - mAXC), 
We have 

mZV+ ml ABC - m L BCD. why? 
Therefore 

mZV+ \mAXC = l 2 mBYD. Why? 

Complete the proof. 



594 Circles and Spheres 



Chapter 13 



31. An angle whose sides are a secant-ray and a tangent-ray from an exterior 
point of a circle is called a secant-tangent angle. Prove that the measure 
of a secant-tangent angle is one-half the difference of the measures of 
the intercepted arcs, 




(Hint: Using the notation of the figure you must prove that 

mZ V = fynBYC - mAXC). 
Use Theorem 13.19 stated in Kxercisc 24. Also see Exercise 3CLJ 

32. An angle whose sides are two tangent-rays from an exterior point of a 
circle is called a tangent-tangent angle. Prove that the measure of a 
tangent-tangent angle is one-half the difference of the measures of the 
intercepted arcs. 




(Hint: Using the notation of the figure you must prove, that 

m£V= limAYB - mAXB) 
Use Theorem 13.19 stated in Exercise 24. Also sec Exercise 30.) 

33. Copy and complete the following theorem which combines the state- 
ments of Theorem 13.20 (see Exercise 30) and Exercises 31 and 32 into 
a single statement. 



13.6 Intercepted Arc*, Inscribed Angles, Angle Miaiure 595 



THEOREM 13.21 Jf an angle has its vertex in the exterior of a circle 
and if its sides consist of two secant-rays, or a secant-ray and a tungeut- 
ray, or two tangent-rays to the circle, then Hie measure of the angle 

34. Complete the proof of the following theorem. 

THEOREM 13.22 The measure of an angle whose vertex is in the 

interior of a circle and whose sides are contained in two secants is one- 
half the stun of the measures of the intercepted arcs. 

(An angle such as the one described in Theorem 13.22 is called a 
chord-chord angle. 




Using the notation of the figure, it is given that S3 and CD are 
secants intersecting at E, a point in the interior of the circle. Prove that 

ml DEB = ^rnDYB + mAXC). 
Exercises 35-44 refer to Figure 13-13. In the figure, VA and VC are secant- 
rays of the circle with center P, VE is a tangent-ray, chords A~D and BC 

intersect at F, and B-P-C. If the measures of arcs AB, BD, £*£, and AC are 
as shown, use the results of Exercises 30, 31, 32, and 34 and other theorems 
proved in this section to find the indicated measure in each exercise. 



35* m£AVG 

36. mCE 

37. mlCVE 

38. mLAVE 

39. mlBFD 

40. mlAFB 
4L mLDEG 

42. ml BEG 

43. ml CAB 

44. mlBEA 




Figure 1*43 



596 Circlet and Spheres Chapter 13 

45, challenge PROBLEM* In the figure, circles C and C with centers P 
and F t respectively, are tangent internally at R and Circle C contains P. 
If RH is any chord (with one endpoint atfi) of C and if H? intersects C 
at Af, prove that M is the midpoint of KS. 




13.7 SEGMENTS OF CHORDS, TANGENTS, AND SECANTS 

If two distinct chords of a circle intersect at an interior point of the 
circle, the point of intersection together with the midpoints of the 
chords determine four distinct segments (other than the chords) which 
are subsets of the chords. For example, in Figure 13-44, chords AB and 
CD intersect at P. The four segments to which we refer are segments 
Ai\ PB, t J", and PD. I 1 is eas> to prove that the protlut 1 \>i 1 he lengths 
of the two segments on AB is equal to the product of the lengths of the 
two segments on CD. 



^- +^ Figure JIM4 

THEOREM 13,23 If two chords of a circle intersect, the product 
of the lengths of the segments of one chord is equal to the product 
of the lengths of the segments of the other. 

Proof: Using the notation of Figure 13-44, we are given a circle with 
chords AB and CD intersecting at P. We are to prove that 

AP-PB = CP-PD, 



13-7 Segment* of Chords, Tangents, and Secants 

We draw AD and EC. Then 



597 



LA=z LC 
£Dm LB. 



(Why?) 

(Why?) 



and 

Therefore 

AADP - ACBP 

by the A, A, Similarity Theorem. It follows that 

(AP, PD) = (CF, PB) 
and that 

AP-PB = CP*PD. 

This completes the proof of Theorem 13,23. 



Figure 13-45 shows a line VT tangent to a circle ut Tand a secant 

FA intersecting the circle in points A and B. in die statements of our 
next three theorems, we refer to segments such as VT, PB, and FA. 




Figure 13-45 



Therefore it is convenient to have a name for each of them. We make 
the following definition. 



Definition 13.18 If Vand T are distinct points and if the 

line VT is tangent to a circle at T, then the segment VT is 

called a tangent-segment from V to the circle. If secant PA 
intersects a circle in points A and B such that A is between P 
and B t then the segment FB is called a secant-segment from 
P to the circle and the segment 1*A is called an external 
secant-segment from P to the circle. 



The proof of our next theorem is Exercise 37 of Exercises 13.3. 



598 Circles and Spheres 



Chapter 13 




THEOREM 13.24 The two distinct tangent-segments to a circle 
with center O from an external point P are congruent and the angle 
whose vertex is P and whose sides contain the I wo tangent- 
segments is bisected by the ray PO. 

Proof: Using the notation of Figure 13-46, given a circle with center 

O, tangent-segments PA and PB, and ray PO, we are to prove that 

J5i fiS FE and that PO bisects LAPB. Give reasons for the statements 
in the proof when asked, 

Z CMP and Z OBP are right 
angles. YVhy? Therefore A GAP 
and AOBP are right triangles, 

OA^OB :Why?) 
and 

FO^FO, 
Therefore 

A OAP ^ A OBP. { Why?) 

It follows that PA Si PB and that ZAPO ^ ZBPO. To complete the 
proof we need to show that O is in the interior of LAPB. 

The union of the given circle and its interior is a convex set. Call 
it S. If Q is any point of S, then OQ < OA, If R is any point of PA, ex- 

4---* 

ccpt A, then OR > OA. Therefore S does not intersect PA except at 
A. Therefore all of S except A lies on one side of PA. In particular, O 
lies on the B-side of PA. Similarly, it may be shown that () lu.-s mi the 
A 'Side of PB. Therefore O is in the interior of Z.APB and since 
Z.APO « ZjBPO, it follows that PO bisects ZAPB. 

17JEOBEM f.3.25 The product of the length of a secai it-segment 
from a given exterior point of a circle and the length of its external 
secant-segment is the same for any secant to the given circle from 

the given exterior point. 



Figure 13-46 



Figure 13-47 




13.7 Segments of Chords, Tangents, and Secants 599 

Using the notation of Figure 13-47, given secant-segments FE and 
FD and external secant-segments J5t and JPG, we are to prove that 
PR • PA = TO ■ PC. Plan: Prove APCB ~ APA D. 

Proof: Assigned as an exercise. 

Our next theorem gives us still another relation between the prod- 
ucts of the lengths of certain segments. Figure 13-48 shows an exterior 
point P of a given circle and a tangent to the circle at C which contains 
the exterior point P. If t is any secant to the given circle which contains 
P and intersects the circle in points A and B, we can prove that the 
product PA * PB is the same as the product PC ■ PC, or (PC) 2 . 




Figure 13-*S 

THEOREM 13.26 Given a tangent-segment PC from P to a circle 
at C and a secant through P intersecting the given circle in points A 
and B, then 

PA • PB = (PC) 2 , 

Proof: Using the notation of Figure 13-48, we are given a tangent- 
segment TC from an exterior point P to the circle at C PB is any secant- 
segment from P and intersecting the given circle in points A and B. We 
are to prove that PA*PB = (PCp. 



L mlPCA = ^mAC 

2. m/,B = Jmulfi 

3. mLPCA = mLB 

4. ZPCA = LB 

5. ZPs ZP 

6. APCA ~ APBC 

7. (PC, PA) = (PB, PC) 

8. PA*PB = (PQ 2 



Benson 

1. The measure of a tangent- 
chord angle is one-half tile 
measure of the intercepted arc. 

2. Why? 

3. Why? 

4. Why? 

5. Why? 

6. Why? 

7. Why? 

8. Why? 



600 Circles and Spheres Chapter 13 

Tt follows from Theorem 13.23 that if P is any point in the interior 
of a circle, the product PA • PB remains unchanged for any chord 7lB 
which contains P. Also, if P is a point in the exterior of a circle as in 
Theorems 13.25 and 1 3.26, the product PA - PB remains unchanged for 
any secant-segment from V and intersecting the circle in points A and 
B, or for any tangent-segment from P to the circle. Theorems 13,25 
and 26 suggest that the value of the "unchanged product" is the square 
of the length of the tangent-segment from P to the circle. This seems to 
have no significance for Theorem 13.23. Figure 13-49 suggests (with 
a bit of help from Pythagoras) that (OP) 2 — r 2 exists even when P is 
inside the circle. As we shall see, the idea of {OP} 2 — r 2 as the value of 
die "unchanged product*' is what relates Theorem 13,23 to Theorems 
13.25 and 13.26. 





Figure 13-49 

Definition 13,19 Given a circle S with center O and radius 
r y and a point P in the same plane as 5, the power of /' with 

respect to S is (OF) 2 — r 2 . 

Given a circle S with center and radius r, a point P coplanar with 
S is in the interior of S, on S, or in the exterior of S, according to 
whether OP < r, OP = r, or OP > i\ hence according to whether 
(OP)'* < r'^iOF} 2 = i* or {OP) z > r 2 , and hence according to whether 
the power of P with respect to S is negative, zero, or positive. If an 
iy-coordinate system is set up in the plane of S with its center O as 
origin, then the power of P with respect to S is 

(OF)* - r* = x 2 + if - r 2 

and 

interior of S = {(*, y) : x 2 + if - r* < 0} 
S = {(*, y) : x 2 + y* - r 2 = 0} 
exterior of S = {(%, y) i x 2 + y 2 — r 2 > 0). 

Compare this with the rep rose utations using set-builder notation in 
Figure 13-9. 



13.7 Segments of Chords, Tangents, ami Secants 601 

We are now ready for the theorem that relates Theorem 13.23 to 
Theorems 13,25 and 13.26. 

THEOREM 13.27 Let a circle S with center O and radius r be 
given. Let P be a point in the plane of 4? and let p be the power of P 
with respect to S. If a line through P intersects S in points A and 
B t then 

1. PA-PB= -p if Pis inside S, and 

2. PA • PB = p if P is on S or outside S. 

Proof; Our proof is by cases. In Cases 1 and 3, EF is the diameter of 
S such that EF contains P, 

Case I. 

PA • PR = FE • PF (Why?) 

= (r - Of)(r + OP) 
= r 2 - (OP) 2 

= -P- 

Casel 




J» = A 



Case 2a. 

PA • PB = ■ PB 
= 
= (Of) 2 - r 2 



Case 2fr. 

PA ■ PB = ■ • 
= 
= (Oi^ 2 - r* 




Case 2a 




Case 2b 



Cow 2c. 

PA-PB = PA • 

■ 

= (OP)2 - r2 

= p. 




^■^ /i 



Case 2c 



602 Circles and Spheres 



Chapttr 13 



Cases 3a and 3b. 

FA-PB = PE-PE (Why?) 
= {OP + r)(OP - r) 
= (Of) 2 - r* 




O.-iao 3b 



The theorems wc have proved in this section enable us to do many 
numerical problems. 

Example 1 In Figure 13-50. CD = 38, CE = 2Q S and BE ^ 24. Find 
DE and AE. 



Solution: DE = CD - CE 

= 38 - 20 m 18. 

Let AE = %; then, by Theorem 13.23, 

*•££ = CE-DE. 
Thus 

*-24 = 20-18 
or 

24* S 360 Figure 1340 

and 

x= 15. 

What is the power of £ in this example? Is the power of E a positive 

number or a negative number? 

Example 2 In Figure 13-51, PB = 70 and PC = 40. Find PA. 




Figure UMQ 




13,7 Sagment* of Chords, Tangents, and Secants 603 
Solution: By Theorem 13.26, (PA) 2 = PB • PC. Therefore 

(PA) 2 = 70-40. 



Thus 



and 



(PA) 2 = 2800, 
(FAV = 400 ■ 7. 



FA = V400-7, 
PA = 20 yT 



What is the power of P in Example 2? Is the power of P a positive 
number or a negative number? 

Exampte 3 In Figure 13-52, FB = 78, AB a 26, and PD = 82. 
Find CD. 




D 
Figure 13-52 

Solution: First we need to find PA and PC. Why? We have 

PA = FB - AS = 78 - 26 = 52, 
By Theorem 13.25, 

PD-PC = PR -PA. 
Therefore 

82-PC=78-52 



PC = 78-52 m 49 19 
82 41 



Thus 



CD = FD - PC a 82 - 49$ = 32|f 
What is the power of F in Example 3? 



604 Circles and Spheres 



Chapter 13 



EXEBCISES 13.7 
I, Prove Theorem 13.25. (liefer to Figure 13-47.) 

Exercises 2-5 refer to Figure 1 3-53 which shows two intersecting chords of 
a circle. 




Figure 13-53 

2L UCE= 7 and DE = 9, find the power of E. 

3. If AB = 24, BE = 8, and C£ = 12, find DE and the power of E. 

4. If CE = -r, DE m 8. AE = 12, and BE = x - 2 f find CE f BE, and the 
power of E. 

5. If AE = x BE = x - 4, CE = 4, and ED = 16 - x, find AE, BE, £D, 
and the power of E. 1 low are A"U and CD related? 

Exercises 6-14 refer to Figure 13-54 which shows two secant-raj's and a 
tangent-ray from an exterior point of a circle. If an answer is an irrational 
number, put it in simplest radical form. (For example, \/32 = 4 \/2 i» sim- 
plest radical form. ) 




Figure 1334 

6. If PE =16,PD= 10, and PB = B t find PC. 

7. Find the power of P in Exercise 6, 

8. Find PA in Kxercise 6, 

9. If ED = 9, DP = 12, and PC m IS, find BC. 



13.7 StgmanU of Chords, Tangents, and Secants 605 



10, Kind the power of F in Exercise 9, 

11, Find PA in Exercise 9, 

12, If PA = 16, FB = 10, and PE = 24, find DE, PC, BC, and the power 
ofP< 

13, If PA m x, PE = 50, PD = 32, and PC = x + 20, find PA, PC, PB, 
and BC, 

14, Find the power of P in Exercise 13, 

15, The figure shows two circles intersect- 
ing at Q and H, a tangent-ray from V 
to the larger circle at P, a tangent-ray 
from V to the smaller circle at S. and 
a lecant-ray from V intersecting the 
two circles in points Q and B. Prove 
that VP = VS. (The line QR in the 
Ggyre is called the radical axis of the 
two circles. It is the set of all points of 
like power with respect to both circles.) 

16, The figure shows two circles tangent 
externally at T and vT is a common 
tangent, A secant-ray from V inter- 
sects the larger circle in points A and 
B, and a secant-ray from V intersects 
the smaller circle in points R and S, 
Prove VA - VB = VR - VS. 




17. Given the figure in Exercise Hi Prove that AVAR ~ AVSB. 

18. Given that the sides of quadrilateral ABCD are tangent to a circle at 
points H, E, /-, P, as shown in the ligurc, prove that 

AB + CD = AD + BC 




B 



606 Circles and Spheres 



Chapter 13 



19. If a common tangent of two circles docs not intersect the segment join- 
ing their centers, it is called a common external tangent If it does in- 
tersect the segment joining, their centers, it is called a common internal 
tangent. In the figure, Kb is a common external tangent of the two 
circles with centers P and Q> PR = 21, QS = 6, and PQ = 25. Find AS. 
(Hint: Draw $T 1 PK at Z) 




20. Cive reasons for steps 2 to 8 in the proof of Theorem 13.26. 

21. Complete the proof of the following theorem. 

THEOREM If A and B are distinct points on a circle and if it is any 
point between A and B, then ti is in the interior of the circle. 

Proof: Set up an xy-eooiduiate system with the origin at the center of 
the given circle and with the X-ftXis the perpendicular bisector of AH 
as shown in the figure. Let r be the radius of the circle. Then there is a 
number a such that -r<a<r and 

A& = {(x, y) : x = a). 




Then the endpoints of AB arc (a, \fF~~- a 2 } and (a, — yr 2 - a 2 }, and 
a point R{ a t tj) is between A and B 

if and only if - \/^ - a 2 < y < V^ - <&, 
if and only if y 2 < r 2 — a 2 . 

Complete the proof hy showing that OR < r and hence that it is in 
the interior of the circle. 



Chapter Summary 607 

22. Let line I be a tangent to a circle ul T, Prove that all points of the circle, 
except 7', are on one side of I in the plane of the circle. {Hint: Let A 
and B be anv two distinct points of the circle such that A *£'£ mid 
B ^= 1\ and suppose that A and B are on opposite sides of I. Then there 
is a point of I between A and B. Why? Therefore I intersects the in- 
terior of the circle. Why? See Exercise 21. Contradiction?) 

23, challenge FROBliKM. Given a right triangle, APQR, with the right 
angle at H, let C bo the circle with center at P and radius PR as shown 
in the figure, Then £Jft is a tangent-segment (Why?) and the circle in- 
tersects QP in two points S and T. Why? Use Theorem 1 3.2fl and prove 
that 

{QFf = (QRf + (RF)Z. 

Thus you will have given another proof of the Pythagorean Theorem. 




CHAPTER SUMMARY 



The following terms and phrases were defined in tills chapter. Be sure 
that yon know the meaning of each of them. 



CIRCLE 

SPHERE 

RADIUS OF CIRCLE (SPHERE) 

DIAMETER OF CIRCLE 

(SPHERE) 
CONCENTRIC CIRCLES 

( SPHERES) 
CHORD OF CIRCLE (SPHERE) 
TANGENT LINE TO CIRCLE 
TANGENT CIRCLES 
TANGENT PLANE TO SPHERE 
SECANT 
CONGRUENT CIRCLES 

(SPHERES) 
INTERIOR OF CIRCLE 

(SPHERE) 



EXTERIOR OF CIRCLE 

(SPHERE) 
GREAT CIRCLE 
CENTRAL ANCLE OF C1RCI £. 
MINOR ARC OF CIRCLE 
MAJOR ARC OF CIRCLE 
SEMICIRCLE 

DEGREE MEASURE OF ARCS 
INSCRIBED ANGLE 
INTERCEPTED ARCS 
CONGRUENT ARCS 
TA NGENT— SEGM ENT 
SECANT— SEGMENT 
EXTERNAL SECANT— 

I GMENT 
POWER OF A POINT 



608 Circles and Spheres Chftptor 13 

There were 27 numbered theorems in this chapter. You should road 
them again and study them so that you understand what they say. The fol- 
lowing are a list of a few of the more important theorems concerning circles. 
You should know the corresponding theorems with regard to spheres where 
applicable. 

THEOREMS 13.3 and 13.4 I ^sl a circle and a line in the same plane 
be given. The line is tangent to the cirde if and only if it is perpen- 
dicular to a radius at the outer end of the radius. 

THEOREM 13.5 A diameter of a circle bisects a chord of the circle 
other than a diameter if and only if it is perpendicular to the chord. 

THEOREM 13,8 Chords of congruent circles are congruent if and 
only if they are equidistant from the centers of the circles, 

THEOREM Hi. 1ft The measure of an inscribed angle is one-half the 
measure of its intercepted arc. 



REVIEW EXERCISES 

In Exercises 1-22, refer to the circle C* with center P shown in Figure 13-55, 
Assume that all points in the figure are where they appear to be ia the plane 
of the circle. Copy the statements, replacing the question marks with words 
or symbols that best name or describe the indicated parts. 




Firuit 13-55 

1. AB is called a \Tj of *h e circle. 

2. PE is called a |T| of the drcle. 

3. AT) is called a [TJ of the circle, KB could also be called a (JJcl the circle. 

4. AB is called a [T]. 

5.IfSnc={T) l then fi? is called a [7] to the circle at '/'. 



Review Exercises 



609 



6. If TK n C = {T}, then TK is to FT at I, 

7, Those points named in the figure and which arc on the circle are jTJ- 

S. Those points named in the figure and which are points in the interior 

of the circle are [?} 
9. [?] is a point in the exterior of the circle. 



10. L ADF is an [7] angle. 

11. L BAD is [?] in arc BAD. 

12. Z ADF intercepts arc |Tj. 

13. DTE is a \J\ arc of the circle. 

14. DAF is a J} arc of the circle. 

15. AM} is a [?J. 

W. mZBAI? is mB7). 



17. L EFT is a [T] angle. 

18. m/-T=[7J. 

19. The power of <? is [TJ. 

20. Hie power of K is g$ 
22. Tlic power of A is \T\ r 
22. A aJ|0 Is similar to AQJ 



In Exercises 23-29 refer to the sphere S with center Q shown in Figure 
13-56, Copy the statements, replacing the question marks with words or 
symbols that best name or describe the indicated parts. 




Figure 13-56 

23. QR is a [3 of the sphere. 

24. If V t Q, Tare collincar, then Vf is a [?] of the sphere. 

25. If W, Q, R are coQmear, then the circle with center Q and containing 
points W t F, R is a of the sphere, 

26. RV is a [Tj of the sphere. 

27. HP is a [JO- 

28. If a n S = {T), then a is Q] to S at T. 

29. If a H S = {T}, then Qf is [7] to a at T, 



9ft Find the radius of a circle if one of its chords 16 in, long is 6 in. from 
the center of the circle. 



610 Circles and Spheret Chapter 13 

31. How far from the center of a circle with radius 16 is a chord whose 
length is 24? 

32. A circle C and a line Hn tin ay-plane are given by 
C = {{ar, y) : a* + y2 = 36} 



(a) Write the coordinates of six points that are on the circle. 

(b) Write the coordinates of three points that are on the line. 

(c) Write the coordinates of two points that are on the circle and on the 
line, 

33. A sphere S in an xys-coordinate system is given by 

S = {{*> y, «) : *• + g* + # m 81}. 

(a) Write the coordinates of eight points that are on the sphere. 

(b) Write the cooreli nates of two points that are in the interior of the 
sphere. 

(c) Write the coordinates of two points that are in the exterior of the 
sphere. 

(d) Write the coordinates of two points on the sphere that are endpoints 
of a diameter of the sphere and are not on any of the three coordi- 
nate axes. 

34. Write an equation of a sphere with center at the origin of an xtp-coor- 
dinate system and which contains the point (4, —2, 5). What is the 
radius of this sphere? 

35. Let a sphere with radius 10 be given. A segment from the center of the 
sphere to a chord and perpendicular to the chord has length 4, Find the 
length of the chord. 

In Exercises 3o 45 refer to die circle with center P shown in Figure 13-57. 
Given the notation of the figure and the degree measures labeled in the 
figure, find the measure asked for in each exercise. You may need to refer 
to the theorems stated in Exercises 24, 30 t 31. 32, and 34 of Exercises 13,6. 




Figure 13*5? 



Review Exercises 



611 



36. mAB 

37. mCT 

38. ml BCD 

39. nUCPT 

40. m£AVC 



41. mlCVT 

42. mlBTG 

43. milBFD 

44. mLABD 

45. ittilP/V 



In Exercises 46-50 refer to the circle with center P shown in Figure 13-58. 

In the figure, VA and VC are secant-rays intersecting the circle in points 

A* B and C, I ), respectively. VT is a tangent to the circle at 71 Chords AD 
and BT intersect at E. 




Figure I3-5S 

40, If VA = 12, VB = 20, and VC = 14, find VD and the power of V. 

47. If VA = 16, and AB = 9, find VT and the power of V, 

48. If AD = 24, AE = 18, and ET m 8, find Brand the power of B. 

40, If BE = 12, ET = x, AE = 18, and ED = x - 2, find x, BT, AD, and 

the power of E. 
50. If BE = 9, EA = x, AE = x + 12, and ED = X - 3, find *, AE, £D, 
and the power of E, 




Chapter 




Brooks /Mankmeyar 



Circumferences 
and Areas 
of Circles 



14.1 INTRODUCTION 

In the first part of this chapter, we consider some of the properties 
of regular polygons that are useful in developing the formula for the 
circumference of a circle and the formula for the area of a circular re- 
gion. We usually say "the area of a circle" as an abbreviation for the 
phrase "the area of a circular region/* or "the area enclosed by a 
circle." 

Tti the last part of the chapter, we depart from our formal geometry 
and present an intuitive approach to the development of the formulas 
for the circumference of a circle, the area of a circle, the length of an 
arc, and the area of a sector. We appeal to your intuition in developing 
these formulas because a formal treatment involves the use of limits, 
a topic that vou would study in a mathematics subject called "calcu- 
lus." We use the idea of a limit to make the formulas seem plausible. 



614 Circumferences and Areas of Clrdts Chapter 14 

14.2 POLYGONS 

In this section we investigate some of the angle measure properties 
of convex polygons. We also consider some of the properties of a cer- 
tain subset of convex polygons called regular polygons. 

Recall the definition of a convex polygon in Chapter 4. We say that 
a polygon is a convex polygon if and only if each of its sides lies on the 
edge of a halfplane which contains all of the polygon except that one 
side. In this chapter all the polygons with which we are concerned are 
convex polygons. Therefore, when we speak of a polygon, we mean u 
convex polygon. 

Recall that two vertices of u polygon that are endpoints of the same 
aide are called consecutive vertices, or adjacent vertices. Two sides of 
a polygon that have a common endpoint are called consecutive sides, 
or adjacent sides. An angle determined by two adjacent sides of a poly- 
gon is called an angle of the polygon. Two angles of a polygon are 
called adjacent angles of the polygon if their vertices are adjacent ver- 
tices of the polygon. 

For the polygon ABCDE shown in Figure 14-1, A and B are ad- 
jacent vertices. AB and AF, are adjacent sides, and Z.A and LB arc 
adjacent angles of the polygon. 
Vertices such as A and C, or such as 
A and D s or such as B and D* and so 
on, are called ncmadjaceiit vertices 
of the polygon. A segment whose 
endpoints are nonadjacent vertices 
of a polygon is called a diagonal of 
the polygon. In Figure 14-1, ~KC is 
a diagonal of the polygon. Name 
three more diagonals of the polygon 
ABCDE. How many distinct diag- 
onals does this polygon have alto- 
gether? How many distinct diagonals are there that have a given vertex 
as an endpoint ? In the work I hat follows we are going to be concerned 
with determining the number of diagonals from an arbitrary vertex of 
a given polygon. Note that a triangle has no diagonals since each pair 
of vertices in a triangle is a pair of adjacent vertices. 

In Chapter 7 we proved that the sum of the measures of the angles 
of a triangle is 180. We then used this theorem to prove that the sum 
of the measures of the angles of a convex quadrilateral is 360. i See The- 
orem 7.33 and its proof.) Let us now review the ideas of this proof W \ 1 1 1 
the aid of the new terminology introduced in the study of areas in 
Chapter 9. 




14.2 Polygoni 615 

If the quadrilateral ABCD shown in Figure 14-2 is a convex quad- 
rilateral, then the diagonal AC parti lions the polygonal region ABCD 
into two triangular regions, ABC and ADC. The union of these two 
triangular regions is the polygonal region ABCD. In the proof of The- 
orem 7.33, we showed that the sum of the measures of the four angles 
of the quadrilateral is the same as the sum of the. measures of a certain 
set of six angles, three from each of the two triangles. In this way we 
obtained 2 « 180, or 360, as the sum of the measures of the angles of a 
convex quadrilateral. 




Now let us extend this idea to convex polygons of 5, 6, 7, 8, or a 
sides, where n is a positive integer greater than 4. Figure 14-3 shows 
pictures of polygons with 5, 6, 7, and 8 sides. The names of these poly- 
gons are pentagon, hexagon, heptagon, and octagon, respectively. 






a 



616 Circumferences and Areas of Circtat Chapter 14 

On a sheet of paper draw four convex polygons, a pentagon, a hex- 
agon, a heptagon, and an octagon, In each polygon, label one vertex A 
and draw all distinct diagonals from A. The diagonals partition each 
polygonal region into a certain number of nonoverlapping triangular 
regions whose union is the given polygonal region. By a procedure sim- 
ilar to the one we used with the quadrilateral, find the sum of the meas- 
ures of the angles of each polygon and summarize the results in a table 
like the one shown in Figure 14-4, Complete the last two rows of the 
table on the basis of your computations for the first five rows. The en- 
tries in the last row shotild l>e formulas involving n. 



Number of 




Number of 


Sum of Measures 


Sides of 


Number of 


Triangular 


of the Angles 


Convex Polygon 


Diagonals from A 


Regions 


of the Polygon 


4 


1 


2 


2 ■ 180 = 360 


5 


2 


m 


m 


6 


3 


m 


m 


7 


4 


m 


m 


8 


m 


m 


m 


9 


m 


m 


m 


n 





m 


m 



Figure 14-4 

On die basis of the results of the computations suggested, it seems 
reasonable to conclude that each convex polygonal region of n sides 
can be partitioned into n — 2 triangular regions by drawing the diag- 
onals from one vertex. Since die sum of the measures of the angles of 
each triangle diat bounds a triangular region is 180, the formula 
(n — 2)180 appears to be a correct formula for determining the sum 
of the measures of the angles of a convex polygon of n sides Wc slate 
this result as a theorem. 

THEOREM 14.1 The sum of the measures of the angles of a con- 
vex polygon of H sides is (n — 2)180. 

Proof: Let A be any vertex of the given convex polygon with n sides 
and let the polygon be A BCD . . . MN as suggested in Figure 14*5. 
Since a diagonal exists from A to each of the n vertices of the polygon 
except the vertices A, B, N (Why?), there are n — 3 diagonals from 
the vertex A. Match A ABC with AC, AACD with AD This es- 



14.2 Polygons 617 



tablishcs a one-to-one correspondence between the set of n — 3 diago- 
nals and the set of all triangular regions with the exception of AMN. 




Figure 14-5 

Therefore there are (n — 3) + 1, or n — 2, triangular regions. The 
union of these n — 2 triangular regions t ABC, ACD, . . . t AMN, is the 
polygonal region 

ABCD , . . \l\\ 

The sum of the measures of all the angles of the triangles that bound 
these triangular regions is (n — 2)180. It follows from the Angle Meas- 
ure Addition Theorem that the sum of the measures of all the angles of 
the triangles 

AAJ5C, AACO t , > . , AAMN 

is the same as the sum of the measures of all the angles of the polygon 
ABCD . . . MN. This completes the proof. 

An important subset of the set of all convex polygons is Lhc set of 
polygons whose sides are all congruent and whose angles are all con- 
gruent. We call such polygons regular polygons. Figure 14-6 shows a 
regular pentagon ABCDE and a regular hexagon ABCDEF. 





618 Circumferences and Areas of Circles 



Chapter 14 



Definition 14 J A regular polygon is a convex polygon ull 
of whose sides are congruent and all of whose angles are 
congruent. 



What do we call a regular polygon of three sides? Of four sides? 
Note that a rhombus (Figure 14-7a) which is not a square has all of its 
sides congruent, but it is not a regular polygon. Why? Similarly, a 
rectangle (Figure 14~7b) which is not a square has all of its angles con- 
gruent, but it is not a regular polygon. Why? 




Figure 1.4-7 



{a) Rhombus 



(b) Rectangle 



Since a polygon of n sides has n vertices and therefore « angles, 
we have an important corollary of Theorem 14.1 that applies to a regu- 
lar polygon of n sides. 

COROLLARY 14.1,1 The measure of each angle of a regular 
polygon of n sides is 

(n - 2)180 
n 

Proof: A regular polygon of fl sides has n angles and each of these 
singles has the same measure as every other angle of the polygon. Since 
the sum of the measures of the angles is (it — 2)180, it follows that each 

angle has a measure of - . 

n 

Tn Chapter 6 we defined an exterior angle of a triangle. We now 
extend the definition to convex polygons of more than three sides. 



Defin ition 1 4.2 Each angle of a convex polygon is called aj i 
interior angle of the polygon. An angle which forms a linear 
pair with an interior angle of a convex polygon is called an 
exterior angle of the polygon. Each exterior angle is said to be 
adjacent to the interior angle with which it Forms a linear 
pain 



14.2 Polygons 619 

Note that there arc two exterior angles at each vertex of a polygon 
as suggested in Figure 14-8 and that, t>eing vertical angles, they are 
congruent to each other. It follows that a polygon of n sides has 2n 
exterior angles. 




Figure 14-8 



Suppose that we are given a convex polygon of n sides and suppose 
that we choose one of the two exterior angles at each vertex. The 
chosen exterior angle and the adjacent interior angle of the polygon 
at that vertex are supplementary. Why? Therefore the sum of their 
measures is 180, Since there arc n vertices, the sum of the measures of 
dl the interior angles and the chosen exterior angles is n • 180* But we 
have shown that the sum of the measures of all the interior angles of 
the polygon is (n — 2)180. Therefore the sum of the measures of all 
the chosen exterior angles (one at each vertex) is 

n(180) - (n - 2)180 = 180« - ISOn + 360 
= 360. 

We have proved the following theorem. 

TTTEOREM 14.2 The sum of the measures of the exterior angles, 
one at each vertex, of a convex polygon of n sides is 360. 

This means that the sum of the measures of the exterior angles of 
a polygon is independent of the number of sides the polygon may have. 
Suppose we consider a particular regular polygon, such as a regular 
hexagon. Wc know that the sum of the measures of all the interior an- 
gles of the hexagon is (n — 2)180 and that the measure of eacli angle 

(n - 2)180 
it 



of a regular hexagon is 



For a hexagon, n = 6. 



620 circumferences and Areas of Clrclt* Chapter 14 

Therefore the measure of each angle of a regular hexagon is 

< 6 ~ 2 > 18 ° = 120. 
6 

It follows that the measure of each exterior angle of a regular hexagon 
is 60 and that the sum of the measures of the exterior angles, one at 
each vertex, is 6" ■ 60 = 360, 

Another way to calculate the measure of each exterior angle of a 
regular hexagon is to use Theorem 14.2. Since supplements of con- 
gruent angles are congruent, It follows that all the exterior angles of a 
regular polygon are congruent. If we choose just one exterior angle at 
each vertex, there are 6 chosen exterior angles of a regular hexagon 
and each of them has a measure of I • 360 = 60, This leads us to the 
following corollary of Theorem \42. 

COROLTARY 14.2.1 live measure of each exterior angle of a 
regular polygon of n sides Is ■ . 

Proof: Assigned as an exercise, 

EXERCISES 14.2 

■ In Exercises 1-6, use the formula of Corollary i.4.1. 1 to find the measure of 
each Interior angle of the indicated regular polygon. 

1. Pentagon A. Octagon 5, 24-gon (24 sides) 

2, Heptagon (7 sides) 4. Decagon (10 sides) 0, ISO-gon 

7, Fi nd t he in ensure of each exterior angle of the regular polygons of Exer- 
cises 1-6 in two differ cut ways, 

■ Copy and complete the statements in Exercises 8 and 9 with the word in- 
creases or the word decreases, 

8» As the number of sides of a regular polygon increases, the measure of 

each interior angle of the polygon [?J. 

&. As the number of sides of a regular polygon increases, the- measure of 
each exterior angle of the polygon [£}■ 

10. The measure of each interior angle of a regular n-gon is 140. Find n. 
{Hint: Solve the equation ( " : i. 2 ) 180 _ 14Q f OT ^j 

11. The measure of each interior angle of a regular n-gon is 150. Find n. 
(See Exercise H>.) 



14.2 Polygon* 621 

12. Can the measure of each interior angle of a regular polygon be 136? 
Explain. 

13. Find the measure of each exterior angle of a regular 12-gon. 

14. The measure of each exterior angle of a regular n-gon is 12, Find n. 

15. Can the measure of each exterior angle of a regular polygon be 50? 
Explain. 

16. Given a regular hexagon ABCDEE as shown in the figure below at left, 
prove that ABDF is equilateral. 





17. The figure above at right shows a regular pentagon ABCDE whose sides 
have been extended to form a five-pointed star. Find the measure of 
each vertex angle of the star (that is, Z P, LQ> LR? L S, L T). 

18. If the skies of a regular hexagon were extended to form a six-pointed 
star, what would be the measure of each vertex angle of the star? 

19. Hie sum of the measures of 14 angles of a polygon of 15 sides is 2184. 
(a) What Is the measure of the remaining angle? 

;b) Could the polygon be a regular polygon? 

(c) Is there enough information to decide whether it is a regular 
polygon? 

20. The sum of the measures of 8 angles of a 9-gon is 1140. 

(a) What is the measure of the remaining angle? 

(b) Could the polygon be a regular polygon? Explain. 

21. Prove Comllary 14.2.1. 

22. Given a regular pentagon ABCDE, prove that diagonal AD is parallel 
to side EC, 

23. Given a regular hexagon ABCDEl', prove that diagonal AD is parallel 
to side BC. 

24. challenge PROBLEM^Given a regular polygon ABCD ... .V of n sides, 
prove that diagonal AD is parallel to side BC if n > 5. 

25. chaij.f.nc;k HKKBJM. Determine the maximum number of acute an- 
gles a convex polygon can have. 



622 Circumferences and Areas of Circle* Chapter 14 

14.3 REGULAR POLYGONS AND CIRCLES 

We biow that three noneol linear points determine exactly one tri- 
angle. That is> given a set of three noneollinear points there is exactly 
one triangle which has these three given points as its vertices. We now 
proceed to show that, given a set of three noncollinear points, there is 
exactly one circle that contains the three given points. In this sense, 
we say that three noncollinear points determine a circle. 

T^t three noncollinear points D, E, F be given as shown in Figure 
14-9, Let of be the unique plane that contains D, E, F. In plane a, let 1% 
be the unique perpendicular bisector of DE and let t-i be the unique 
perpendicular bisector of FE, Let P be die unique point of intersection 
of ti and Z 2 - (How do wc know that k intersects Z 2 ?) Then P is equidis- 
tant from D and E because it lies on the perpendicular bisector of IM> 
and P is equidistant from E arid F because it lies on the perpendicular 
bisector of EF. Therefore P is equidistant from D, E, and F. It follows 
that P is the center of a circle C with radius 

r = PD = PE = PF. 



hf-.^ 




Figure 14-9 

Therefore C is a circle which contains D i E x F* Furthermore, C is the 
only circle which contains D t £, F. For if C were any other circle with 
center F and radius r* and containing points D, F., F in plane a, then 

¥ = FD = FE=FF 

and F would be equidistant from D, E, and F, Therefore F would lie 
on h and 1% the perpendicular bisectors of DF and EF in plane a. Since 
there is only one point that lies on both of these lines, F = P> ¥ = r, 
and hence C = C. Therefore there is exactly one circle which contains 
any given set of three noncollinear points. It follows that there is 
exactly one circle that contains the vertices of any given triangle, 
We have proved Theorem 14.3. 



14.3 Regular Polygons and Circles 623 



Definition 14,3 The circle which contains the three vertices 

of a given triangle is called the circumscribed circle, or cir- 
cumcircle, of the triangle arid we say that it circumscribes 
the triangle. The triangle is said to be inscribed in the circle 
and is called an inscribed triangle of the circle. 



THEOREM 14.3 A given triangle has exactly one dreumeirde. 

We now extend the statement of Theorem 14.3 to Include any reg- 
ular polygon, Wc want to prove that, given any regular polygon, there 
is exactly one circle that contains all the vertices of the polygon (that 
is. the given regular polygon has exactly one circumcircle). 

THEOREM 14.4 A given regular polygon has exactly one 
circumcircle. 

Proof: Let a regular polygon of n 
sides be given. (In Figure 14-10, we 
have shown a polygon of 6* sides.) Let 
Q be the unique circle with center P 
and radius r which contains A, B, 
and (7. (How do we know this circle 
exists and is unique?) We shall prove 
that PD = r and hence that D lies on 
circle Q, A similar argument could be 
given to show that each of the verti- 
ces of the given polygon lies on Q, 

Figure 14>10 




1. PA = PB = PC = r 

2. Z2s Z3 

3. mLABC = m/BCD 

4. mZ2 = m/.$ 

5. mZl = mZ4 

6. Z 1 == 14 

7. AB S DC 

8. PB m PC 

9. &PBAj= APCD 

10, PA B PD 

11. PD= PA = r 



Reason 

1. Definition of circumcircle 

2. Why? 

3. Why? 

4. Why? 

5. Angle Measure Addition 
Theorem (3, 4) 

6. Why? 

7. Why? 

8. Why? 

9. S.A.S, Postulate {7, 6, 8) 

10, Why? 

11. Statements JO and 1 



Thus D is on Q and this completes the proof for point D, 



624 



Circumf»r»nc«s *nd Artai of Clrcltt 



Chapter 14 



Definition 14.4 {See Figure 14-11.) The rlmrffimfWllir of a 
regular polygon is the center of its circumscribed circle. A 
circumradius of a regular polygon is a segment (or its length) 
Joining the center of the polygon to one of the vertices of the 
polygon. An inradius of a regular polygon is a segment ior its 
length) whose endpoints are the center of the polygon and 
the foot of the perpendicular from the center of the polygon 
to a side of the polygon, A central angle of a regular polygon 
is an angjle whose vertex is at the center of the polygon and 
whose sides contain adjacent vertices of the polygon. 




figure I Ml 



Some authors use radius Instead of circumradius, and apotfiem 
instead of in radius. We prefer the more descriptive terms* circumradius 
and hirudins. They are the radius of the circumscribed circle and the 
radius of die inscril>ed circle of a regular polygon. We know that a 
regular polygon has a circumscribed circle (Theorem 14.4) and we shall 
see that a regular polygon has an inscritwd circle (Theorem 14.7). 

Note that, in connection with a regular polygon, each of the words 
circumradius and inradius is used in two different ways: ( 1 ) as one of a 
set of segments and (2) as a positive numl>er. For example, the inradius 
of a regular polygon means the number that is the length of a segment 
— any one of the segments defined as an inradius in Definition 14.4. 
We shall sec in Theorem 14.8 that all these segments have the same 
length. On the other hand, an inradius of a regular polygon is a seg- 
ment, usually one of the segments defined as an inradius in Definition 
114, but it might also mean any radius of its inscribed circle. Similarly, 
the circumradius of a regular polygon means a number, whereas a 
circumradius of a regular polygon usually means any one of the seg- 
ments joining the center of the polygon to a vertex of the polygon, but 
it might also refer to any radius of die circuinscril>ed circle. The con- 
text in which the word is used should make it easy to decide which 
meaning is intended. 

It follows from the definition of a central angle that a regular poly- 
gon of n sides has n central angles. 



14.3 Regular Polygons and Circles 625 

Suppose that we are given a regular polygon ABCD . . . MN of n 

sides. (Figure 14-12 shows a regular polygon of 6 sides.) Let F be the 
circunicentcr of the given polygon. Il follows that all of the triangles, 
APAB, AFBC, A PCD, , . , , APJVA, are congruent by the S.S.S. 

Postulate. 




Kipire 14-12 



(Prove that A PAS = A PUG) Let us agree to call each of these tri- 
angles (that is, a triangle whose vertices are the center and the end- 
points of a side of the polygon) a central triangle of the regular poly- 
gon. It follows from the definition of congruent triangles that all of the 
central angles of a given regular polygon are congruent. We combine 
these results into the statement of our next theorem. 



TI7EOREM 14.5 Let a regular polygon of n sides be given. Then 
all the central triangles of the gjven polygon are congruent and all 
of the central angles of the given polygon are congruent. 

Again, suppose that we are given a regular polygon ABCD . . . MN 
of n sides. (Figure 14-13 shows a regular polygon of 6 sides.) 




Figure 14-13 



Let P be die cii'cumcenter of the given polygon and let PR, PS, FT, » . « » 
PVbe the n inradii of the polygon. By the definition of an inradius, the 



626 Circumferences and Area* 0* Circle* Chapter 14 

segments PR, M, PT, . . . , PV are the altitudes to the bases IB, BC, CD, 
. « < , RA, respectively, of the central triangles, APAB, APBC, APCD, 
. , . , APNA, of tlic given polygon, By Theorem 14.5 these central tri- 
angles are all congruent. Since corresponding altitudes of congruent 
triangles arc congruent, it follows that all the inradii of a given regular 
polygon are congruent This means that, in the plane of the given poly- 
gon, the points R, $, T Vile on a circle {) whose center is P and 

whose radius, a, is the inradius (a number) of the polygon. Since 
PR I AB^PS^ SC^ff A OR . . . t FT A KE, it follows that each of 
the sides AB, BC, CD* . . , , NA of the given regular polygon is tangent 

to the circle Q and that the points R, S, T, , V are their respective 

points of tangeney. 

Definition 14.5 A circle is said to be inscribed in a polygon 
and is called an inscribed circle, or iiicircle, of the polygon if 
each of the sides of the polygon is tangent to the circle. We 
also say that the polygon circumscribes the circle. The center 
of an incircle of a polygon is an incenler of the polygon* 

We now show that there is exactly one circle inscribed in a given 
regular polygon. Let a regular polygon such as the one in Figure 14-13 
be given. 

We have demonstrated that the circle Q with center P and radius a 
is an inscribed circle of the given polygon. It can be shown that a point 
which is equidistant from the sides of a regular polygon is also equi- 
distant from its vertices and hence must be the circumeenter. Since 
there is only one circumeenter, it follows that there is only nine ineen- 
ter. Therefore Q is the only inscribed circle of the polygon. We have 
proved the following two theorems. 

THEOREM 14.6 All the inradii of a given regular polygon are 
congruent 

THEOREM 14.7 There is exactly one circle that is inscribed in 
a given regular polygon. 

We can now think of the center of a regular polygon as its incenter 
or its circumcenter. 

Recall from Chapter 10 that two polygons are similar if there is a 
correspondence between their vertices such that corresponding angles 
are congruent and lengths of corresponding sides are proportional. It 



14.3 FUgular Polygons and Circles 627 

follows from the definition of similar polygons that if two regular 
polygons are similar, then they have the same number of sides, Is the 

converse statement true? That is, is it true diat if two regular polygons 
have the same number of skies, then they are similar? This brings ns 

to our next theorem, 

THEORFM 14,8 Two regular polygons are similar if they have 
the same number of sides. 



Proof: Let two regular polygons ABCD , . , M .V and 
A'B'C'D' , . » M'N\ each having n sides, be given as suggested in Fig- 






Figure 14-14 

ure 14-14 We want to prove that the correspondence 
ABCD . . . MN * — * A'B'Ciy . . . 1TW 

is a similarity. By the definition of regular polygon and Corollary 14. LI, 

we have 



m 



n 



and 

mLA* s m£W = m£C = mlD' = • • • = mlN' 

_ (n - 2)180 
n 

Therefore 

mLA = m£A\m£Sm mlB' mlN = mlN', 

and the corresponding angles of the two polygons are congruent. 



628 Circumferences and Areas of Circles Chapter 14 

It follows from the definition of regular polygon that there are two 
positive numbers -9 and $' such that 

AB = BC-CD = •> =NA = s 

and 

AW a B'C = CD' m , , . = N'A' = s'. 
Then 

AB = s = -£■•• = 4-A'F, BC = * = 4**' ■ 4-B^ etc. 

It follows that 

(AB, BC. CD, , . , , NA) = (A'B\ B'C, CD\ , . ♦ , iVA ) 

with proportionality constant k = -^ , Therefore the lengths of the cor- 
responding sides of the two given polygons are proportional and 

ABCD ...MN~- A'&Cfy . - . M'N'. 
This completes the proof of Theorem 14,8. 

THEOREM 14.9 The perimeters of two regular polygons, with 
the same numl>er of sides, are proportional to the lengths of their 
circiunradii or their in radii. 

Proof: Let two regular polygons ABCD . . . MN and A'B'Ciy . . . 
M f N* t each having n sides, be given as suggested in Figure 14-15. 




A H B 

Figure 14-15 




Let p and p', s and $', t and r\ a and a' be their respective perimeters., 
lengths of sides, circumradii, and inradii. Let P and W be the centers of 



14,3 Regular Polygons and Circles 629 

the two polygons and consider the two central triangles 

AAPB of ARCD . . , MN 
at id 

AA'FB' of A'B'CD' , , . Af.V. 
Since 



ml ABB = mlATB' = 

n 

(you will be asked to show this in the Exercises), it follows that 
AAFB — AA'FB' by the S.A.S. Similarity Theorem, Therefore 

(AB.r,a) = (A r B' f r>,tri. Why? 

Since corresponding altitudes of similar triangles are proportional to 
the lengths of any two corresponding sides, we have 

($, r, a) = {s\ f, a'}. 

Finally, 

p = ns for ABCD . . , MX and p' = ns' for A'B'CD' . . . M'.Y. 

Since 

ps' = (ns)s' = (ns> = pfs, 

it follows that 

and hence 

(p. s, r» a) =: (p\ *', /, a'). 
This completes the proof of Theorem 14.9* 

Since all central triangles of a given regular polygon of n sides are 
congruent, and since they partition the given polygonal region Into n 
nonoverlapping triangular regions, it is easy to prove that the area of 
a regular polygon is equal to one-half the product of its inradius and 
perimeter. We state this as our next theorem, 

TITEORFM 14 JO The area of a regular polygon is equal to one- 
half the product of its inradius and perimeter, that is, S = -jflp. 

Froof: Assigned as an exercise. 



630 Circumference* and Aft*« vt Clrclei Chapter 14 

The last theorem of this section follows easily from Theorems 14.9 
and 1410, 

THEOREM 14,11 The ureas of two regular polygons with the 
same number of sides are proportional to the squares of their 
circum radii (or the squares of their in radii, or the squares of their 
side lengths, or the squares of their perimeters). 

Proof: Assigned as an exercise. 

As we said in the introduction to this chapter, the idea of a limit is 
important in developing formulas for the circumference and the area 
of a circle. We conclude this section with a brief discussion of se- 
quences and limits. 

An infinite sequence of numbers, denoted by {%„}, is a sequence 
*i» *fc *& • • • , *■» • * « in which, for every positive integer n, *„ is a 
number. In some applications it is convenient to start counting with 
some integer other than 1 . Thus £3, jt|, *a, . . . , x n> . , . is an infinite 
sequence. 

Example 1 Let the nth term of a sequence be given by x„ = 2n. 
Write the first five terms and the 40th term of the sequence {2n}, 

Solution; 

*! = 2 • 1 = 2. 
*a = 2*2 = 4. 

* 3 == 2*3 = 6, 
x, = 2 ♦ 4 = 8. 
* 5 = 2-5 = 10. 
xw = 2 ■ 40 = 80. 

Therefore die first five terms of the sequence are 2, 4, 6, 8, 10 and the 
40th term is 80, 

Example 2 Let the nth term of a sequence be given by 

1* + 1 



*» = 



n 

Write the first five terms, the J 00th term, and the 1000th term of the 



sequence 



m 



Solution: 





1 


+ 


1 






*1 = 




1 




— 


2. 


*2 = 


2 


+ 
2 


1 


— 


3 
2 


*3 = 


3 


+ 


1 




4 




3 






3 




4 


+ 


1 




5 


X4 = 




4 




= 


4 



14,3 Regular Polygons and Circles 631 



5 + 1 6 

*S = = — = ~ * 

5 5 

IftA . 100+1 101 

100 " 100 " 100 
1000 + 1 1001 

* ,000= 1ooo~-iooo 



In Example 2, as n gets larger and larger, the terms of the sequence 
gj$f closer and closer to some particular number. What number is it? 

In Example 1, as n gets larger and larger, do the terms of the 
sequence {2n} get closer and closer to some particular number? If 
so, what number is it? 

If the terms of a sequence {x n } get arbitrarily close (as close as we 
desire) to a particular number L as n gets larger and larger, we then 
say that the nth term of the sequence is approaching L (denoted by 
x n —*L) and we call L the limit of the sequence. Thus, in Example 2 S 

n 

and we call 1 the limit of the sequence | 1 , In Example 1, the 

sequence {2n} has no limit Find the limit (if any) of the sequence — 1 . 

EXERCISES 14.3 

L It follows from Definition 14.4 that a regular polygon of n sides has n 

central angles. Prove that the measure of each central angle of a regular 

360 
polygon is — — . 

2. Prove that each central triangle of a regular polygon is an isosceles 
triangle. 

3. Prove that each central triangle of a regular hexagon is an equilateral 
triangle. 

4. Prove that the length of a side of a regular hexagon is equal to the 
circumradius of the circuiucirclc. 



632 Circumferences and Area* of Circlt* Chapter 14 

5. Prove thai an in radius of a regular polygon bisects a side of the polygon 
and hence lies on the perpendicular bisector (in the plane of the poly- 
gon) of the side of the polygon, 

6. Prove that the bisector rays of the interior angles of a regular polygon 
are concurrent at the center of the polygon. 

7. Prove that the measure of a central angle of a regular polygon is equal 
to the measure of an exterior angle of Lhe polygon, 

8. Two regular pentagons have sides of length 4 in. and 5 in,,, respectively . 
What is the ratio of the perimeter of the smaller pentagon to the 
perimeter of the larger one? What is the ratio of their eircurnradii? Of 
their inradii? Of their areas? 

9. The area of a regular 12-gon is *£■ times the area of a second regular 
12-gon. What is the ratio of their perimeters'? Of their eircuinradii? Of 
their inradii? 

10. A regular hexagon has twice the area of another regular hexagon. What 
is the ratio of the perimeter of the smaller hexagon to the perimeter of 
the larger one? What is the ratio of the lengths of their sides? Of their 
cirenmradii? 

11. Two regular polygons of the same number of sides have perimeters of 
36 in. and 48 in., respectively. The in radius of the first polygon is 
3\/5hx What is the hirudin* of the second polygon? What is the area 
of each of the polygons? 

12. Find the circum radius, inradius, and area of an equilateral triangle 
each of whose sides is of length 2 \/3. 

13. Find the circum radius, hirudins, and area of a regular hexagon each of 
whose sides is of length 12. 

14. Show that the inradhis of a regular hexagon is — ^ — . s, where s is the 

length of a side of the hexagon, 

15. Derive a formula for the urea S of a regular hexagon in terms of the 
length s of its side. (Hint: See Exercise 14.) 

16. Use the formula you derived in Exercise 15 to find the area of a regular 
hexagon each of whose sides is of length 12. Does your answer for die 
area agree with that of Exercise 13? 

17. If a regular hexagon and a regular triangle are inscribed in the same 
circle, prove that tire length of the side of the Hexagon is twice Lhe ui- 
radius of the triangle, 

18. A square is inscribed in a circle of radius 1, A second square is circum- 
scribed about the same circle. Find the area and the perimeter of each 
square, 

19. Repeat Exercise 18 using regular hexagons instead of squares. 

20. The length of each side erf a regular hexagon is 8 s/3. Find the area of 
the hexagon in two different ways. 



14,3 Regular Polygon* and Circfct 633 

21* The circum radius of a regular pentagon is r and the length of each of 
its sides is s. Find the area of the pentagon in terms of r and s. 

22. In Exercise 21, if the eircumradius of a second regular pentagon is 2r, 
find die area of the second pentagon in terms of t and *. 

23. Prove Theorem 14 JO, 

24. Prove Theorem 14 J L 

25. Write reasons for statements (2), (3), (4), (6), (7), (8), and (10) In the 
proof of Theorem 14,4. 

26. What is the 1000th term of the sequence {-}? What is the millionth 

term? Which of these two terms is closer to zero? Could we make — as 

n 

dose to zero as we desire by choosing n large enough? 

27. Find the first five terms, the 100th term, and the 1000th term of the 

28. What is the millionth term of the sequence 1 ^ 1 of Exercise 27? 

Find (he limit (if any} of the sequence. 

In Exercises 29^37, a formula for x m is given. Find *i, x», x& and Xi<>, What 
is the limit (if any) of the sequence («■}? 

■fc^-l + i M.^iiJOO 1 + 50 

~2n 2u 2 tt 

2n n 

* L *-* + 5T- 2n m *---2nT- 

32.^=2- y? tXm = !£jLL 

33, ^ = J- 
2" 

38. Let a circle be given. For each n,n > 3, let a regular polygon of n sides 
be inscribed in the circle. Let »„ be the perimeter of the n-gon. Does 
Px, p<» Ps» • ♦ ■ , p«» ■ • • define a sequence? Do you think {p„ } has a limit? 
If so, how would you describe the limit? 

39. 1 ,et a circle be given. For each n, n > 3, let a regular polygon of n sides 
be inscribed in the circle. I .et S n be the area of the polygon. Does S3. S*. 
Sa, . , , , S nw . , . define a sequence? Do you think {S„} has a limit? If so, 
how would yon describe the limit 

"J^- . — -jT — J has a limit. 

What is it? 



634 Circumferences and Are« of Circle* 



Chapter 14 



14.4 THE CIRCUMFERENCE OF A CIRCLE 

Thus far, in our formal geometry, length has not been defined for 
anything except segments. If a path from one point to a second point is 
such that every point of the path lies on the same segment, then the 
length of that path is, of course, the length of the segment joining the 
two points. However, If the path is a circular arc, what is the distance 
from the first point to the second point along the circular arc, that is, 
what is the length of the arc? The degree measure of the arc would 
not be a satisfactory way of describing its length since it is possible 
for two arcs to have the same degree measure and Lo have different 
lengths as suggested in Figure 14-16. 




Figure M-I6 

Each of die arcs AB and A B' shown in the figure has a degree 

measure of 90. But it certainly seems reasonable to think of the arc AB 

as having a greater length than the arc A B\ Wc start by explaining 
what we mean by the length of circular arcs and then deriving ways 
of finding such lengths. We first proceed informally, referring to the 
physical world. 

You may have been asked, at one time or another in yonr study of 
informal geometry, to wrap a string around a circular object, then pull 
it out straight, and measure its length, liy doing this you are able to 
arrive at an approximation to the distance around the object However, 
we cannot describe the process of wrapping a string around a circular 
object in our formal geometry. 

We call the '"distance around a circle" the circumference of the 
circle and denote it by C A more sophisticated approach to finding an 
approximation to the circumference of a dido is in terms of the perim- 
eLers of regular polygons inscribed in the circle. There was no difficulty 



14.4 Th« arcumftmtc* of • Circle 635 

in defining the perimeter of a polygon because the sides of a polygon 
are segments and each of these segments has a length. But a circle con- 
tains no segment of a line, and thus we cannot define its circumference 
(or perimeter) so simply. It seems reasonable lo suppose that if we want 
to find the circumference of a circle approximately, we can do it by 
inscribing in the circle a regular polygon with a large number of sides 
and then measuring or computing the perimeter of the polygon* 

Given a circle, let »„ be the perimeter of a regular polygon of n sides 
inscribed in the circle. Then as n gets larger and larger, the number p„ 
increases, that is, each term in the sequence {p„} ts greater than the 
preceding term. For example, we can inscribe a square in a circle. By 
bisecting the central angles of the square we obtain a regular octagon 
inscribed in the same circle as shown in Figure 14-17. Using the Tri- 
angle Inequality Theorem, it is easy to show that the perimeter p$ of 
the regular octagon is greater than the perimeter p 4 of the square. If we 
continue to bisect the central angles, wc obtain a regular 16-gon, a reg- 
ular 32-gon, and so on. The perimeters of these regular polygons p 4 . ptt* 
Pig* P32. . - ■ form an infinite sequence of numbers, and each term in 
the sequence is greater than the preceding term. 




Hfttra 14-17 

It is shown in the calculus that if a sequence of numl>ers is increas- 
ing (that is, if each term in the sequence is greater than the preceding 
term), and if the sequence is bounded (that is, if there is a number that 
is equal to or greater than any term in the sequence)* then the sequence 
has a limit. The squcnoe p*» pg, pm, ps*jt • ■ ■ described above in con- 
nection with the circle shown in Figure 14-17 can be shown to be 
bounded. In fact, it can be shown that any square that circumscribes 
the given circle has a perimeter that is greater than any of the terms in 
the sequence. 



636 Cfreumfcrancet and Areas of Circlai Chapter 14 

Let a sequence of perimeters of regular polygons of n sides in- 
scribed in a given circle be denoted by {p«}. Let die limit of this se- 
quence be denoted by C, that is, Hm p n = C. We are now ready for 
our formal definition of circumference as die limit of the p, . 

Definition 14.6 The circumference of a circle is the limit 
of the sequence of perimeters p a of the inscribed regular pol- 
ygons (that is, C = lim p n ). 

Note that we are forced to use limits in defining the circumference 
of a circle. In order to derive the formulas for the circumference and 
for the area of a circle* we need a theorem about limits which we state 
here without proof. We can treat the theorem as a postulate, although 
it is not a postulate concerning our formal geometry. Rather t it is a 
postulate concerning the real number system. 

THEOREM 14.12 Let {**} and {y n } be two sequences of real 
munhers with nth terms x n and y n , respectively, 

1 . If the limit of x« is Lj and the limit of y n is I^ t then the sequence 
whose nth term is x„y n has a limit, and 

ton x n y n = Li'Lz. 

2. If die limit of *„ is Lj and the limit of y H is L^ J* 0, then the 
sequence whose nth term is — has a limit and lim — = — -. 

3. If x n = y„ for every positive integer n > I and if {*„) and { y„] 
each has a Limit, then lim x» = lim i/„. 

4- If k is a real number and if x n = k t for every n > l f then 

lim x n — k. 

Example 1 If lim ^^=A = 2 and lim -^-- = 3, then 
n n + 2 



rnl 



U,n(fczLa._aL) = 2-3 = 8 
\ n n + 2/ 

\ n n -f- 2/ 3 

2 Find lim x n if x?, = 2 for every » > 1. 



14.4 Tht Cfrcumferancs of a Circle 637 

Solution; If x, { = 2 tor every n > 1, then {ar rt } is a sequence of num- 
bers whose every term is 2; that is, {a&,} = 2, 2, 2 By part 4 of 

Theorem 14.12, lim x n = 2, 

Before we derive a formula for the circumference of a circle, we 

need to know that the number ~ , where C is the circumference of a 

a 

circle and d is its diameter, is the same for all circles, That the num- 
ber ~ is the same for all circles is a corollary of our next theorem. 



THEOREM 14.13 If C and C arc the circumferences of any two 
circles with diameters d and il\ respectively, then 

(C, d) = (C. <f). 

Proof: Let K t K' be any two circles with circumferences C. C and 
radii r, r', respectively, as shown in Figure 14-18. 





Flf»U-Ifi 

Let [p H ] he the sequence of perimeters of regular polygons of n 
sides inscribed in circle K with radius r and let [p^] be the sequence of 
perimeters of regular polygons of n sides inscribed in circle K* with 
radius r\ By Theorem 14.9, 

(p», r> f (pi />. 

It follows that 

hence that 

(1) (Pm <*) = (Pn, <#% 

where d and ci are the diameters of the circles K and K', respectively. 



638 Circumferences and Artai of Ckrd«i ChapUr 14 

By Definition 14.6, 

lim p n = C 

and 

Ump^C, 
It follows from Equation (I) that 

(2) £L = iL. 

By part 3 of Theorem 14.12, we get from Equation (2) 
(3) lim &■ = lim 4- 

By part 1 of Theorem 14.12, 

lim £=■ = £, 

and, since d and d' are the same for all n, it follows from part 4 of The- 
orem 14.12 that 

hm- = -. 
Substituting these last two results in Equation (3), we have that 

SL-± 

C d' 

or that 

and the proof is complete. 



COROLLARY 14.13.1 If C and d are the circumference and di- 
ameter, respectively, of a circle, then the number ^ is the same for 

a 

all circles. 

Proof: Let K, K' be any two circles with circumferences C, C and 
diameters d, d\ respectively, By Theorem 14.13, we know that 

(C, d) - «7, d'\ 



14.4 The Circumference of a Circle 639 
By alternation, we get 

(C, C) = (d, d'). 

Therefore -y = ■*£-, This proves that the number -r is the Same for any 

two circles. 



Definition 14.7 If C is the ctreum fcrence of a circle and d is 

its diameter, then the number -4, which is the same for all 

a 

circles, is denoted by the Greek letter r. 

It follows from Corollary 14,131 and Definition 14,7 that 

C = ml 

for any circle with circumference C and diameter d. Since d = 2r, 
where r is the radius of the circle with diameter d t we have 

C = 2*rr 

as another formula for the circumference of a circle. 

It has been proved that it is not a rational number, that is, v cannot 

be represented by -^ where a and b are integers with b =fi 0. However, 
h 

we can approximate tt as closely as we desire by means of rational num- 
bers. Some of the more common rational number approximations to sr 
are 3.14, 2 ^, and 3.1 416. It has been shown that ff, to ten decimal 
places, is 3.1415926535. 

EXERCISES 14.4 

1. Show that |f{ is a closer approximation to w than is 2 ? 2 

2. Show that 2 f z: 3,14 to the nearest hundredth. (We read "^" as *'is 

approximately equal to.") 

In Exercises 3-10, C, r t and d represent the circumference, radius, and di- 
ameter, respectively, of a given circle. Express answers in exact form, in 
terms of it if necessary. 

3. If r = 7, find a 

4. If C = 83«r, find ft 

5. If d =12.5, find C. 



640 Circumferences and Ar»n of Circle* Chapter 14 

6. If f = 36.4 in., find C. 

7. If C = 36.4 in., find r. 
& If C = 2**, find r. 

9, If r = 3ir, find C. 
10. If C= 14*26*, find Id 

1L In Exercise 6, use v = -4^- and find C to tlie nearest inch. 
12* In Kxerdse 6, use % = 3.14 and find C to the nearest inch. Compare 
your answer with that of Exercise 11. 

13. In Exercise 7, use r, = -3^- and find r to the nearest hundredth of an 
inch. 

14. In Exercise 7, use *r = 3.14 and find r to the nearest hundredth of an 
inch. Compare your answer with that of Exercise 13. 

15. Prove that the circumferences of two circles are proportional to their 
radii, 

10. The circumference of one circle b two-thirds the circumference of a 
second circle. What is the ratio of the radius of Oie first circle to the 
radius of the second circle? 

17. Two circles have radii of 15 and 23, What is the ratio of the circum- 
ference of the smaller circle to the circumference of the larger deck? 
What is the ratio of the diameters of die two circles? 

18. A lire on a wheel of a car has a diameter of 28 in. 1 f the wheel makes 12 
revolutions per second, wliat is the approximate speed of the car in 
miles per hour? (Use w == 4j^-.) 

19. Given the same car as in Exercise 18, how many revolutions per second 
would the wheel make if the car were traveling 50 miles per hour? (Use 
tr= 2 7 2 -) 

20. A square ABCD fa inscribed in a circle with center E as shown in the 
figure. If the radius of the circle is 7, find the perimeter of the square. 
(Hint: Show that &EAB is an isosceles right triangle.) 




14.4 The Circumference of a Circle 641 

21. A square ABCD is inscribed in a circle with center £, Another square 
PQHS is circumscribed about the same circle as shown in the figure. If 
the radius of the given circle is 14 in., find the following: 

(a) The perimeter of each square to the nearest inch. 

(b) The area of each square to the nearest square inch, 

(c) The circumference of the circle to the nearest inch. (Use v = Q) 




v 
E 






B 



22. The perimeter of a square is equal to the circumference of a circle. If 
the radius of the circle is 1, find the area of the square in terms of ?■ 

23. Show that if the radius of a circle is increased by 1 unit, the circumfer- 
ence of the circle is increased by 2*r units. 

24. Show that if the circumference of a circle is increased by 1 unit, the 

radius of the circle is increased by «=r« units. 

7 2-ff 

25. Assume that the surface of the earth is u sphere. Imagine that a steel 
band is fit snugly around iho equator (a great circle of the earth";. Sup- 
pose that one foot is added to the length of the band and that it is raised 
a uniform amount all the way around the earth. To the nearest inch. 
how many inches will the new band be above the earth? (See Exercise 
24.) 

26. Assume that the surface of an orange is a sphere. Imagine that a steel 
band is placed around a great circle of the orange so that it just fits. Sup- 
pose that one foot is added to the length of the hui id and that it is raised 
a uniform amount all the way around the orange. To (lie nearest inch, 
how many inches will the new bund Ixs above tlic orange? (See Exer- 
cises 24 and 25.) 



642 Circumference* and Ar»« of C4rol»» 



Chapter 14 



Exercises 27-36 refer to Figure 14-19. ABCD is a square inscrilwd in the 
circle with center P, The bisector rays of the four central angles of square 
ABCD intersect the circle in points E t h\ G» H t respectively. The polygon 
AEBFCCDH is a regular octagon inscribed in the circle, and TQRS is a 
square circumscribed about the circle with points of tangcucy E, F, C, H. 
The radius of the circle is 8. If an answer to an exercise is an irrational nunt- 
her, give a rational approximation to the nearest tenth. 




Figure l+iu 



27. Find the length * of a side of square ABCD, 

28. Find the perimeter of square ABCD. 

29. Find the area of square ABCD, 

30. Find the length ** of a side of the regular octagon AEBFCCDH. 

31. Find the perimeter of the octagon. 

32. Find the area of the octagon. 

33. Find the length s" of a side of the square TQRS. 
M. Find the jwrimeter of square TQRS. 

35. Find the area of square TQRS. 

36. Find the circumference of the circle. 

37. Imagine an infinite number of regular polygons inscribed in a circle 
with radius r. The first polygon has 3 sides, die second polygon has 
4 sides, the third has 5 sides, and so on. For every n, n > 3. let p„, a n , ft* 
and S, he the perimeter, iriradius. side length, and area, respectively, 
of the regular inscribed polygon with n sides. What is lim p m ? lim a£ 
I in i *»? lim S m ? 



14.5 Area* of Circle*. Arc Length: 5ectw of a Circle 643 



14.5 AREAS OF CIRCLES; ARC LENGTH; SECTOR OF A CIRCLE 

In Chapter 9 we considered areas of polygonal regions. Recall that 
a polygonal region is the union of a polygon and its interior. In this sec- 
tion we are concerned with areas of circular regions. We make the 
following definition. 



Definition 14.8 A circular region is the union of a circle and 
its interior. 



As we said at the beginning of the chapter, "the area of a circle" is 
m abbreviation for "the area of a circular region," or for "the area 
enclosed by a circle." 

We now proceed to get a formula for the area of a circle. We 
already have a formula for the area of a regular polygon of n sides 
which is 

where a n is the inradius of the polygon, p„ is the perimeter of the 
polygon, and S„ is the area of the polygon. 

If P„ is a regular polygon of n sides inscritied in a circle with center 
Q and radius r t as shown in Figure 14-20 (in the figure, n = 8), we 
observe that the area of the inscribed polygon is less than the area of 
the circle. 




Figure 14-20 



644 Circumf irenc** and Aran of Clrcltt Chapter 14 

For the expression 

where n = 3, 4, 5 t . . . , there are three sequences involved; 

{S n }, {a n ) t and {p n }. 
t*et us consider each of these sequences separately, 

1. The sequence {*>„}. As noted above, for each n, S„ is always less 
than S, the area of the circle, The difference between S s and _S 
can be made arbitrarily small by talcing n large enougli. It seems 
reasonable then to say that 

lim S» = S. 

Definilon 14.9 The urea of a circle is the limit of the se- 
quence of areas of the inscribed regular polygons. 

Thus lim S„ = S, by definition. 

2. The sequence [a„). Since the length of a leg of a right triangle 
is less than the length of the hypotenuse of the right triangle, we 
observe that the inradfus <*, is always less than the radius r of 
the circle for each particular value of n, However, the difference 
between a H and r can be made arbitrarily small by choosing n 
laige enough- Thus it seems reasonable to say that 

lim On = r, 

and we accept this feet without proof. 

3. The sequence {«„}, By the definition of the circumference C 
of a circle, 

lim p n = C, 
Now we have 

S n = fop*. 
By parts 3 and 4 of Theorem 14.12, we get 

(1) lim S„ = ^lim a n p n > 



14.5 Atms of gmlM, Are Ungth; Sector of a Circle 645 
By definition 14.9, 
2 : lira S» = S. 

By part J of Theorem 14.12, we get 

(3) lim anpn = lim o„ ■ lim p n 
But 

(4) lim (in = r 
and 

(5) lim p rt = C. 

Substituting the results of (2), (3), (4), and (5) in Equation (1), 
we obtain 

(6) S = {tC 

as a formula for the area of a circle. Since 

C = 2*rr, 
we get by substitution into Equation (6), 

S = ^r-2irr 
or 



S = 

as a formula for the area of a circle— a formula that should be 
familiar to all of you. 

We now state this result formally as a theorem. 

THEOREM 14.14 The area S of a circle with radius r is wr 2 , that 
is, 

COROLLAR Y 14.14, 1 The areas of two circles are proportional 
to the squares of their radii. 

Proof; Assigned as an exercise. 

We have defined the circumference of a circle to be the limit of the 
sequence of the perimeters of the inscribed regular polygons. We now 
proceed to define the length of an arc of a circle as a certain limit. 



646 Circumftnncti and Areas of Circfti Chapter 14 

Consider an arc AM of a circle with center V as shown in Figure 

14-21. For k > 2, let P u Pa, »%..., Ph-l be points on Ait such that 
each of the & angles 

ZAVP b ZP i VP 2 , / P*VP 3 , . .. , Z ft., VB 
has a measure of ~ m^C3. (in Figure 14-21, k = 4.) 




Figure 14-21 

Let 

A* = AP t + PjJPa + f»p, + . . . + P*iR 
Tiros, in Figure 14-21, 

A 4 = APi + Pjft + P^ + P 3 B. 
If we bisect each of the central angles 

£AVP U L PiVP 2 LPh-iVB, 

we obtain k more points Q u fy Q k onAB such their 

A 2 * = AQ t + QiP x + P^ 2 -f Q, £ p 2 + . . . + Q fc R 
In Figure 14-21, 
As = (A0, + Q l P 1 ) + (P^g + £*P 2 ) + (P 2 £ 3 + ^) 3 p 3 ) 

+ (p*q 4 + g^. 

Now 

APi + Pi^i > APi, 

F,p 2 + <?aP 2 > -PiPa, 

Pap3 4- Ws > PaP* 

^Q* + Q 4 B > P3H, Why? 



14.5 Arras of Circles; Arc Langth; Sector of • CJrcl* 647 

It follows that As > A*. If we continue to bisect the central angles 
with vertex V, we obtain a sequence of sums 

A*, As, Aje, . . •» A„ 

(where n = k, 2k t 4k t &k t ,* .) in which each term in the sequence is 
greater than the preceding term. Also, this is a bounded sequence. 
Each number in it is less than the circumference of the circle. There- 
fore the sequence {A n } has a limit and we define that limit to be the 

length of arc A B, 

Definition MAO The length of arc AB (denoted by I A B) is 
the limit of [A m ] where 

A„ == APi + P X P 2 + • • - + Pn-iB 

and where Pi, P> £ F n -i are n - 1 distinct points of A& 

subtending congruent angles at the center V of the circle 

containing AH. 

We now have two types of measure for arcs of a circle: their de- 
gree measures and their lengths. 

Definition 14 J I In the same circle or in congruent circles, 
two arcs arc congruent if and only if they have the same 
length. 

Thus, if arcs AB and &W are arcs of congruent circles and if 
AB 5b A'B\ we have 

(1) IA3 = IA T &'* 
Also, from Definition 1347, if AB s X" # B\ we have 

(2) mAB = mA%. 
Hence, if A& and A^T are congruent arcs of congnient circles, then 

(JAB, life) = {mAB, mAW). 

lu other words, lengths of congruent arcs of congruent circles are pro- 
portional to their degree measures. This is a trivial assertion, of course. 

Suppose, for example, that IAB = 100 and mA$ = 20. What we are 
inserting is that 

(100, 100) = (20, 20). 



648 Circumferences and Ar*i« o» Circtoi 



Chapter 14 



It is also true (not trivial and \vc shall not prove it here) that the lengths 
of any two arcs of congruent circles are proportional to their degree 
measures. We state this as our next theorem. 

THEOREM 14,15 The lengths of arc* of congruent circles are 
proportional to their degree measures. 

Thus, if K and K' {as shown in Figure 14-22) are congruent circles 
and if A& is an arc of K and A'B' is an arc of K\ we have 

(IAB> U%) = (mAB t mAW). 

Suppose that A'B' in Figure 14-22 is a semicircle. Then 
niAW = 180 and LYB' = *r. Why? 





Let b\B 1ms denoted by L and mAB be denoted by M. It follows from 
Theorem 14.15 that 



Therefore 
and 



(/,, vr) = (M, 180). 
180L = tttM 



We have proved the following theorem, 

THEOREM 14,16 The length L of an arc of degree measure M 
contained in a circle with radius r is uiLWr, that is, 

\ 1 g!U/ 



I 



\180/ 



14.5 Area* of Circle*; Arc Untfh; Sactw of a Circle W9 

Example 1 Find the length of an arc of a circle with radius 12 if the 
degree measure of the arc is 60. 

Solution; Use the formula of Theorem 14.16 with M = 80 and 
f = 12, Therefore 

Exam pit- 2 Find the degree measure of an are of a circle with radius 

2 if the length of the arc is 4? ■ 

•i 

Solution: Solving the formula given in Theorem 14,16 for M, we 

M _ (J80U (Show this.) 

to 
3 



We are given that L = 4?- and that r = 2. Therefore 



M-J3L&- 120. 
sr«2 3 

Note that if M = 360 in the formula of Theorem 14.16, we get 

Thus L equals the circumference of the circle as it should. 

Figure 14-23 shows a portion R of a circular region which is 

hounded by two radii, EA, FB, and an arc AB of the circle. We call H 
a sector of the circle. A more precise definition follows. 




Hgurc 14-23 



650 Circumferences and Areas of Circles 



Chapter 14 



Definition 14.12 (See Figure 14-24.) Given a circle of ra- 
dius r with center P and an arc AB of this circle, the union of 
all segments T<$ such that Q is a point on arc AB is called a 

sector. We call AB the arc of the sector and we call r the 
radius of the sector. 



Figure 14-24 




Suppose that we arc given u sector of a circle with radius r and 

center V as shown in Figure 14-25. Let AB 1*? the arc of the sector. 

For n > 2, let P h i*2, - • . P»_i be n - 1 points on SB such that each 
of the n angles 

lAVP lt £P 1 VP*... 1 £P m _ l VB 
has a measure of — • mAB, (In Figure 14-25, n = 4.) 



Figure 14-25 



A 






""^T^J" 






i y\ 




r 




\p* 







14.5 Are** of Cbcitft; Arc Ungth; Sector of a Orcle 651 

Let 

(1) An = APi + Fa?* + ■ • + Pn-iB. 

Let S., be the area of the polygonal region VAP1P2 • •*!* l®l th en 

(2) Sn = 4fl a (AJ>i) + J*p!A) + • ; ■ 4- i««(P n iB), 

where a n is the altitude to each of the bases 3JV F^T\ P*-iR ©f 

triangles AAVPi, AP1VP2, • • • , AP n _iV7J, respectively. From Equa- 
tion (2) we get 

(3) % = ^niAPi + PjPa + ■ .. + P„-iR); 
hence 

(4) S H = fynA*. 

by substitution from Equation (1) into Equation (3). 

We can make the difference between S„ and S (the urea of the 
sector) arbitrarily small by choosing n large enough. Similarly, the dif- 
ference between a n and r can be made arbitrarily small by choosing n 
large enough. Tt seems reasonable, then, that 

Urn S n = S and lim o„ = r. 

By Definition 14.10, lim A n = L, the length of arc Afr By applying 
parts 3 and 4 of Theorem 14.12 to Equation {4) t we obtain 

(5) lim S„ = Jiim 0*3,, 

From equation (5), it follows that S = \tL is a formula for die area of 
a sector with radius r and arc length h, We state this result as our next 

theorem. 

THEOREM 14. J 7 The area S of a sector is one-half the product 
of its radius t and the length L of its arc, that is, 

S = |«L 

If we combine the results of Theorems 14.16 and 14.17, we get the 
following theorem. 

THEOREM 14,18 If M is the degree measure of the arc of a sector 

with radius r, then the area «S of the sector is U==s } aw®* that is, 

\,jou/ 

Proof: Assigned as an exercise. 



€52 Circumferences and Areas of Circlet Chapter 14 

Example 3 Find the area of a sector with radius 10 if the degree meas- 
ure of the arc of die sector is 72, 

Solution: If we use the formula of Theorem 14.1 8 with r = 10 and 
M = 72, we obtain 

Note that, in the formula of Theorem 14.17, if L is the circumfer- 
ence C of the circle, then 

L = %<nr 

in id 

S = ^r • 2frr = wr 2 , 

as it should for a circle, Similarly, in the formula of Theorem 14.18, if 
the given arc is the circumference of the circle, then M = 360 and 

S = (Jlftn* = ««, 

as it should for a circle. 

EXERCISES 14.5 

■ In working title exercises of this set, do not use an approximation for w un- 
less instructed to do SO. 

L Prove Corollary 14.14.1. Let Sj and Sv be the areas of two circles with 
radii r t and r%, respectively. Prove that 

A. Si) ? fe*, *& 

2. The area of one circle is \ times the area of a second circle. What is the 
ratio of the radius of the first circle to the radius of the second circle? 

3. The radii of two circles are r and 2r, How does the area of the larger 
circle compare with the urea of the smaller circle? How does the cir- 
cumference of the larger circle compare with the circumference of the 
smaller circle? 

4. Show that if r < 2, the area of a circle is less than the circumference of 
the circle. (Of course, the urea units and the length units are in different 
systems. If the length units are inches and the area units arc square 
inches, and if r < 2, then the statement to be proved asserts that the 
number of square inches in the area is less than the number of inches in 
the circumference.) 



14,5 Aran of Circl**: Arc Ltngth; Stctor of a Clrcla 653 
In Exercises 5-11, r is the radius of a circle and S is its ana. 

5. If r = 8, find & 9. If r = y/4\, find S. 

6. If S = \mv, find r 10. If r = x/TTir. find S, 

7. If S = 154, find r. U. If S = 75.30, find r. 

8. If r = 3ir, find S. 

12. In Exercise 7, use ir - If- and find r, 

13. In Excrdsc 9, use ?r = 3.14 and find S to the nearest tenth. 

14. hi Exercise 11. use w = 3.14 and find r to tlie nearest tenth. 

In Exercises 15-18, the area of a drab is given. In each exercise, find die 
circumference of the given circle, 

15. 81*r 17. 81 

16. 49w 18. 49 

In Exercises 19-22, the circumference of a circle Is gfven. In each exercise, 
find the area of the given circle. 

19. 40*r 21. 40 

20. lfor 22. 16 

23. The figure shows two concentric circles with radii 7 cm, and 10 cm. The 
union of the two circles and the shaded portion between them is some- 
times called an annul us. Find the area of the annulus. 




24. A square and a circle have the same area. 

(a) Find the perimeter of tlie square in terms of the radius r of the circle 

(b) Find the circumference of the circle In term* of the length i of the 
side of the square. 

25. Find the ureas of the inscribed and curcuinscribed circles of a square 
with side length 12. 

26. In Exercise 25, find die area of the annulus {see Exercise 23) between 
the two circles. 



654 Circumferences and Areas of Circle* Chapter 14 

27. The radius of the circle with center C is 24. The radius of the circle 
with center C is 12, If mX3=60 and mA'fi'=60 T show lAB^ZlAll'. 

.A 





28. For the circles of Exercise 27, if IAB = lA'B' and mAB = 60, find 
mA'B'. 

Given a circle with radius r and an arc AB of the circle, use the given infor- 
mation in Exercises 29-35 to find the indicated measure. 

29. If r = 12 and mAB = 45. find IAB, 

30. If r = 1 and IAB = ^- t find mAB. 

4 

31. If /AB = 11^ and mAJI = 60, find r. 

6 

32. If LAB = ^L and mA3 = 270, find r. 

33. If r = 18 and mAB = 240, find /AB. 

34. Kr=9 and mAB = 132, find IAB, 

35. If r = 4 and fAH = Git, find iriAB. 



36. Given a circle witli radius fa 1, find the degree measure of each of the 
following arcs of the circle. 



a) AB tf IAB = £ 

' 6 

(b) .4C if IA 1; = £ 

4 

(c) ADiflAD = Z 

(d) AE if iH = J 

(e) AF if 1AF = |£ 

3 

t AC if 1AC - $f 

4 

(g) ah if zaT/ = ^ 

6 
(h) A/if 7A/ = ff 



(i) A/ifJAJ = ~ 



o-- 



;j) AK if lAK = ^ 



(k) Af,ifZAL = 



4c 



AM if 1AM = 



(m) AJV if IAN = 2f 

(n) /IP if LA? = ^ 
(o) AQ if &® = ll£ 



14,5 Areas of Clrct««i Arc Length; Sector of a Circle 65S 

37. Prove Theorem 14,18, (Hint: Combine the results of Theorems J 4, 16 
and 14.17.) 

38. The radius of a circle is 12 in, Find the area of a sector with the follow- 
ing arc length: 

(ft) 4r; (b) 2.7* (c) 8w (d) -?- 

39. The radius of a circle is 15 in. Find the area of a sector with an arc whose 
degree measure is 

(a) 60 (b) 144 (c) 1 (d) 330 

40. Let a circle with center P and radius r be given. Points A and B arc 
points of the circle such thai m Z A PB = 120 and the area of sector A PB 

is 12ir. Find r and /AB. 
41* In an xy-plaue, let sets C and I be defined us follows: 

C = {(.t-, y) : x* + t/ a ■ 64}, 
I = {{x, u):ij = x). 

Let P be the point where / intersects C in the first quadrant, let O be the 
origin of the given xy-plane, and let B he the point cm the positive uc-axis 
where C intersects the x-axis. Find the area of the sector POB. 

Exercises 42-44 refer to Figure 14-26 which shows a circle with center V* 
and chord AB. The shaded portion hounded by the chord AB and the arc 

AB is called a segment of the circle. l.et h he llw altitude to AB of A AVB, 

let AB = s t let VA = r = VB, and let L - 1A&. 




Figure 14-26 

42. challenge problem. Derive a formula for h in terms of r and .?. 

43. challenge problem. Derive a formula for the area S of the segment 

of the circle in terms of r, $ t and L 

44. See Exercise 43. Find S if r = 5, s = fl, and L = ~. (Use -n = 3.14 
and round your answer to the nearest tenth.) 



656 Circumferences and Areas of Circlet 



Chapter 14 



45, challenge problkm. The figure shows three congruent circles with 
centers P, Q, R. Each of the circles is tangent to the other two and the 
points of tangency are A, B„ C, 




(a) Show llutt AP{)R is equilateral. 

(b) If the radius of each of the circles is 12, find the area S of the shaded 

region bounded by arcs AB t BC, and AC. (Use it = 3.14 and round 
your answer to the nearest tenth.) 



CHAPTER SUMMARY 

In this chapter we defined the following terms and phrases. Be sure that 
you know the meaning of each of them. 



INTERIOR ANGLE OF A 

POLYGON 
EXTERIOR ANGLE OF A 

POLYGON 
REGULAR POLYGON 
( JrU I \IS( H IRED CIRCLE 

(CIRCUMCLRCLE) 
INSCRIBED POLYGON 
INSCRIBED CIRCLE 

(INCIRCLE) 
CIRCUMSCRIBED POLYGON 
CENTER OF REGULAR 

POLYGON 
CIRCU M RADIUS OF 

REGULAR POLYGON 



INRAD1US OF REGULAR 

POLYGON 
CENTRAL ANCLE OF 

REGULAH POLYGON 
CENTRAL TRIANGLE OF 

REGULAR POLYGON 
CIRCUMFERENCE OF 

CIRCLE 
THE NUMBER v 
CIRCULAR REGION 
AREA OF CIRCLE 
LENGTHS OF ARC 
SECTOR 

ARC OF A SECTOR 
RADIUS OF A SECTOR 



Chapttr Summary 657 

There were 18 theorems in tliis chapter, most of which consisted of 
formulas. In developing many of these formulas, we used the idea of a 
LIMIT of a sequence of numbers. A complete treatment of limits is too 
difficult for a first course in geometry. Our goal was to give an intuitive 
feeling for the concept <jf a limit and to make the formulas seem plausible, 
Be sure that you know the following list of formulas and that you know how- 
to apply them. 

Sum S of the measures oftiie interior angles of a convex polygon ofn 
sides: 

S s (n - 2)180, 
Measure m of each angle of a regular polygon of n sides: 

, , ft - 2 > 18Q , 
n 

Sum S of the measures of the exterior angles, one at eacli vertex, of a 

convex polygon of n sides: 

S = 360, 

Measure m of each exterior angle of a regular polygon of n sides: 

^360 

ft 

Perimeter of a regular polygon of n sides: 

p = 11% 

where p is tlw perimeter and s is the length of a side of the polygon. 

Area of a regular polygon of n sides: 

S = iap, 
whew S is Hie area, a is the inradhis, and p is the perimeter of the polygon. 

Circumference of a circle: 

C = vdorC = 2m, 
where V is the circumference, r is tlte radius, and d it the diameter of the 
circle. 

Area of a circle: 

where S is the area and r is the radius of the circle. 
Length of an arc of a circle: 

where L is the length of tlw arc, M i* the degree measure of the arc, and r 
is the radius of the circle. 

Area of a sector: 

S = ±rL, 

where Sis the area, r is the radius, and L is the length of die arc of the sector. 



ess Circumferences and Area* of Clrcln Chapter 14 

REVIEW EXERCISES 

1. Find the measure of each interior angle of a regular pentagon; of a 
regular 7-gon. 

2. Find the measure of each exterior angle of a regular 15-gon. 

3. If the measure of each interior angle of a regular n-gon fa 150, find n. 

4. If the measure of each exterior angle of a regular n-gon ii 40 t find n. 
5* If the measure of each of the n central angles of a regular n-gon is 20, 

find n. 
6- Using only a pair of compasses for drawing circles and a straightedge 

for drawing lines, explain how yon would construct a regular hexagon; 

an equilateral triangle? a square. 
7. Find the perimeter and the area of a regular hexagon each of whose 

sides is 20 cm, long. 



In Kxercises 8-13, the radius of a circle is given. In each exercise, find the 
area and the circumference of the circle. Use tt = 3.14 and express each 
answer to the nearest tenth, 

8. r = 13.5 cm. 

9, r= 1 

10, r = 6.04 in. 

11. r= 3,14 ft, 

12. r = 12 

13, r = 62.8 



In Exercises 14-19, the circumference Oof a circle or the area S of a circle 
is given. In each exercise, find the radius of the circle. Give an exael answer 
in each case. 

14. S = 64w 

15. C = 64* 

16. S = 216V sq. cm. 

17. S = 225 sq. in. 

18. C = 14&7 ft 

19. C = 72 cm. 



In Exercises £0-24, r is the radius of a circle, L is the length of an arc of the 
circle, and M is the degree uieusure of the arc. 

20. If r = 16 and M = 80, find L. 

21, If r = 4 and L = 3ir t find hi 



Rtvlvw Extrclttt 



659 



22. If L = -^ and M = 240. find r. 

23. If r = 10 and L = i^, find M. 

24. ff r a 45 and M m -220, find L. 

25. An annulus (the shaded region shown in the figure) has an inner radius 
of x and an outer radius of x + 3, If its area is 48?r. find its inner and 
outer radii. 




Iii the figure, AC?, BC. fixU, and ffiB are semicircles, with A, U. C 
collinear. If AC = BC = 7, find the area of the shaded region. 

fa..-?.) 




£7. Find the area of a sector if the radius is 14 and if the length of the arc 
of the sector is lit;. 



660 Circumferences and Areas of Circles Chapter 14 

28, In the figure, /\ABC is ft right triangle with the right angle at C. An, 

BV, and AC arc semicircles wilh diameters AB t BC : and AC", respec- 
tively. If a, b, c arc the lengths of the sides EC, KTL AB, respectively, 
show that the area of the semicircle with diameter c is equal to the sum 
of the areas of the two semicircles with diameters a and b* 




29. In the figure, A ABC is a right triangle with the right angje at C, KB is 

a diameter of circle JfC, and AKC and BYC arc semicircles with diam- 
eters AC and BC. respectively. Show that the sum of the areas of the 
two shaded regions is equal to the area of the triangle- 




Review Exercises 661 



30. CBHJUJKN&B raoBX.EM. In the figure, A BCD is a square euch of whose 
sides is 10 cm. long, P, Q t R t S are the midpoints of sides A~R, fi€, CD, 

DA, respectively. Arcs EF, FC, Oil, HE are arcs of circles with centers 
P, Q, R, S, respectively, and are tangent to the diagonals of square 
ABCD at points E, P, G, If. 




(a) Prove that PQRS is a square. 

(b) Find die area of the shaded region bounded by the four arcs EF, 

FG t Gtf , and HE. 




Vim Bucher/Fhoto Researchers 



Areas 

and Volumes 

of Solids 



15.1 INTRODUCTION 

In your Study of informal geometry you may have learned formulas 
for finding surface areas and volumes of some of the familiar solids. 
In this chapter we review and extend this phase of geometry- We con- 
tinue our somewhat informal development of geometry from Chapters 
13 and 14. A formal development of these formulas belongs properly in 
a calculus course. The development of this chapter is designed to make 
the formulas plausible. Emphasis is placed on understanding the for- 
mulas and on using them. 

In Chapter 13 we studied circles and spheres, A sphere may be 
thought of as a solid or as a surface. In this book a sphere is a surface, 
and tlie union of a sphere and its interior is a spherical region. In- 
formally speaking, the area of a sphere is a number that expresses the 
measure of the sphere, and the volume of a sphere is a number that ex- 
presses the measure of the associated spherical region. 

Although prisms, pyramids, cylinders, and cones may be thought of 
as either solids or surfaces, in this book we consider them as solids. The 



664 



Areas and Volumes of Solids 



Chapter 15 



area of a cylinder (more properly., the surface area of a cylinder), for 
example, is a number that expresses the measure of the surface of the 
cylinder, and the volume of a cylinder is a number that expresses the 
measure of the cylinder itself, 

15.2 PRISMS 

Figure 15-1 shows some diagrams of prisms. We might think of a 
prism as the solid swept out by a polygonal region moving parallel to 
itself from one position to anodier. Each point P in the region moves 
along a segment PF as suggested in Figure 15-2, and all of these seg- 
ments are parallel to each other. The prism is the union of all such seg- 
ments. Wc make these ideas formal with the following definition. 








Figure 15-1 



15.2 Prisms 



.-■53 



Definition 15.1 ;'See Figure 15-2.) Let a and ft be distinct 
parallel planes. Let Q and Q* be points in a and ft, respec- 
tively. Let R be a polygonal region in a. For each point P in 
R let F be the point in ft such that PF || QQ*. The union of 
all such segments PF is a prism, if QQ' is perpendicular to 
a and ft, die prism is a right prism. 




Figure 15-4 



In this chapter we shall limit our discussion of prisms to those 
prisms whose bases are convex polygonal regions, that is f regions whose 
boundaries are convex polygons* 



Definition 15,2 (See Figure 15-2.) Let R r be the polygonal 
region consisting of all the points F in ft. The polygonal re- 
gions R and R' arc called the bases of the prism, Depending 
upon the orientation of the prism it is sometimes convenient 
to call one of the bases the lower base and the other base the 
upper base. Sometimes we call the lower base simply the 
base. A segment that is perpendicular to both a and ft and 
with its endpoints in these planes is an altitude of the 
prism. Sometimes the length of an altitude is called the alti- 
tude of the prism. 



Prisms are often classified according to their bases. Thus a triangu- 
lar prism is one whose base is a triangular region; a rectangular prism is 
one whose base is a rectangular region, and so on. 



65ft 



Ar#av and Volumes of Solids 



Chapter 15 



Figure 15-3 shows a triangular prism. The bases are the triangular 
regions ABC and A'B'C. The triangular region DEF in the figure, 
which lies in a plane parallel to the plane of the base, is called a crass 
section of the prism. 



\- 




Figure 15-3 



Definition 15.3 If a plane parallel to the plane of the base 
of a prism intersects the prism, the intersection is en' led a 
cross section of the prism. 



Is the triangular region ABC a cross section of the prism shown in 
Figure 15-3? Is the region A'B'C a cross section of the prism? 

We say that a point P of the lower base corresponds to a point R of 
a cross section (other than the lower base) if PR , Qty. We call A DEF 
in Figure 15-3 the boundary of the triangular region DEF. Similarly. 
A ABC is the boundary of the lower base of the prism, & A'B'C is the 
boundary of the upper base, and so on. 

THEOREM I5>] The boundary of each cross section of a tri- 
angular prism is congruent to the boundary of the base of die prism. 

Proof: Let the triangular region ABC in plane a be the lower base of 
the prism and the triangular region A'B'C in plane j8 be the upper base 
as shown in Figure 15-3. Let y be a plane parallel to a and intersecting 
AA', BB' t CO in points D % E, F, respectively. Then, by definition, the 
region DEF is a cross section of the prism. That the region DEF is a 



15.2 Prism* 667 

triangular region can be proved using separation properties. We omit 
the details here. We shall prove ADEF « A ABC. 

If y = a, then D = A, E ! = H. F = C, and ADEF = AABC. 
Therefore ADEF s AAiiC. Why? Suppose, then, that y=£aas sug- 
gested In Figure !5-3. Bv the definition of a prism, there are points Q 
in a and Q'Jn p such that A A' \ QQ' and BB' \ QQ'. Then AA' | BB' 
and AD || BE. Also, by Theorem 8.12, Aft I DE. Therefore ABED is 
a parallelogram and 0E == AB. In the same way we can prove 
W z. AC and FE Sk CB. Therefore A DEF s AABC by the S. S. S. 
Postulate. Since y is an arbitrary plane parallel to a and intersecting 
the prism, we have proved that the boundary of each cross section of a 
triangular prism is congruent to the boundary of its base, and the proof 
is complete. 

Since the upper base of a prism is a cross section of the prism, we 
have the following corollary. 

COROLLARY 15.1.1 The boundaries of the upper and lower 
bases of a triangular prism are congruent 

THEOREM 15.2 (The Prism Crow Section Theorem) /Ml cross 
sections of a prism have the same area, 

Proof: IM the prism as shown in Figure 15-4 be given. (We have 
shown a prism whose base is a polygonal region consisting of five sides. 
The following argument can be modified to apply to a prism whose 
base is a polygonal region consisting of n sides, where n > 3.) Let It be 
the base and let R' be a cross section of the 
prism. Then R can be divided into nonover- 
lapping triangular regions 6), £2, *3 as shown 
in Figure 15-4, Let t' lt & g be the corre- 
sponding triangular regions in B\ Then the 
boundary of h is congruent to the boundary 
of fj, the boundary of h is congruent to the 
boundary of t 2 , and the boundary of 1 3 is con- 
gruent to the boundary of £3, Why? The areas 
of t[, &, t'3 are equal, respectively, to the areas 
0* h, h, k- Why? The area of R f is the sum 
of the areas of r[, if, tk, and the area of R is 

the sum of the areas of t-u tz, h- Why? Since these two sums are equal, 
it follows that the area of R' is the same as the area of R. Since R' is an 
arbitrary cross section of the prism, it follows that all cross sections 
have the same area and die proof is complete. 




668 Areas and Volumes of Solids Chapter IS 

COHOLLARY 15.2.1, The two bases of a prism have die same 



an. -a, 



Proof: Assigned as an exercise. 



Definition 15.4 (See Figure 15-5.) A lateral edge of a prism 
is a segment AA'. where A is a vertex of the base and A' Is the 
corresponding vertex of the upper base. Given any side of 
one base of a prism, the lateral face of the prism correspond- 
ing to that side is the union of all segments FF parallel to a 
lateral edge and with P on the given side of die base. The 
lateral surface of a prism is die union of its lateral faces. The 
total surface of a prism is the union of its lateral surface and 
its bases. 




Hgu» 15-5 

For tlie prism shown in Figure 15-5, AA' is a lateral edge and 
AHB'A' is a lateral face. Name four other lateral edges and four other 
lateral faces in the figure. 

Note that die lateral edges of a prism are parallel to the segment 
QQ' in the definition of a prism. The segment AB in Figure 15-5 is not 
a lateral edge of the prism although AB is an edge of the prism. Indeed, 
it is an edge, or side, of a base of the prism. 

THEOREM 15,3 The lateral faces of a prism are parallelogram 
regions and the lateral faces of a right prism are rectangular regions. 

Fmvfi (See Figure 15-5.) A complete proof involves a discussion of 
separation properties which we omit here. Suppose that we are given a 



15.2 Prisms 669 



prism with points labeled as in Figure ,15-5. We shall content ourselves 
with proving that ABB'A' is a parallelogram and that ABB' A' is a rec- 
tangle if the prism is a right prism. Similar arguments could he given for 
each of the lateral faces of the prism. 

By the definition of a prism, there is a segment (5p' such that 
AA' || QQ' and M'|| QQ' Therefore Id' \\ BB'. Why? By the defini- 
tion of a prism, the planes containing the bases are parallel. Therefore 
it follows from Theorem 8.12 that AH' || AB. This proves that i\BB'A' 
is a parallelogram. 

If die prism is a right prism, then ^?' is perpendicular to plane a. 
Why? Therefore AA'' a (Why?) and AA"' I AB. Why? It follows 
that ABB' A' is a rectangle. 



Definition 15.5 A parallelepiped is a prism whose base is 
a parallelogram region. A rectangular parallelepiped is a right 
prism whose base is a rectangular region. A cube is a rec- 
tangular parallelepiped all of whose edges are eongruenL A 
diagonal of a parallelepiped is a segment joining any two of 
its vertices which are not contained in the same lateral face 
or tiase of the parallelepiped. 



Figure 15-6 shows pictures of a parallelepiped, a rectangular par- 
allelepiped, and a cube* In each picture, the segment rlS is a diagonal 
of the parallelepiped. How many diagonals does a parallelepiped have? 



.'.,!„ 



Rrc'-juifiuUr 
Parallelepiped purillatfpspftd 

an" 




*■ 



-' 



X 



1 II 

_J \ 



Figure lo-fi 



Definition 15.6 The lateral surface area of a prism is the 

sum of the areas of its lateral faces. The total surface area of 
a prism is the sum of die lateral surface area and the areas of 
die two bases. 



670 Areas and Volumes of Solids 



Chapter 15 



EXERCISES 15.2 

1. Copy and complete; 

All the faces of a parallelepiped (lateral, 
upper base, and lower base) are [?] 
regions. 

2. Copy and complete: 

Alt the faces of a [JJ parallelepiped are 
rectangular regions, 

3. Copy and complete: 

All the faces of a [JJ are square regions. 

Exercises 4-16 refer to the prism shown in Figure 15-7. In this figure. Pis a 
point in a and V is a point in /? such thai. PF JL a. In each exercise, com- 
plete the statement. 




4. The region ABCDE is called a [JJ of die prism. 

5. The region A'lfCtXE' h caBod u [JJ of the prism. 
G. BB' is called a (an) |JJ of the prism. 

7. There are [Jj lateral edges in all. 

8. Counting the lateral edges and the edges (sides) of the two bases t there 
are [jj edges in all. 

9. The parallelogram region BCCB' is called a (art) [JJ of the prism, 
10. There are [JJ lateral faces in all. 



15,2 Print* 671 

It, Counting the lateral faces and the two liases, there arc [T| feces in all 

12. A is called a (an) [T) of the prism. 

13. There are (TJ vertices in all- 

14. If V is the number of vertices, £ is the numher of edges, and F is the 
number of faces, then V — £ -j- F = [T], 

15. PP is called an |T| of the prism. 

16. If M' _L 0. then the prism is called a {?}. 

17. Prove that the total surface area of a cube is 6e 2 , where e is the length 
of one of its edges. 

IS. Fmd the total surface area of a cube whose edge is 7 cm. in length. 

19. Prove that the total surface area of a rectangular parallelepiped is 

lab + 2lic + 2uc, 

where a and h are the dimensions of the base and c is the altitude of the 
prism. Draw an appropriate figure, 

20. Find the total surface area of a rectangular parallelepiped if the dimen- 
sions of the base are 4 cm. by 6" an. and if the altitude of the prism is 
8 cm, 

21. Given that the pentagonal prism of Figure 15-7 is a right prism, that the 
lengths of the edges of the base are 3, 7, 4, 9 1 and 6, and that the alti- 
tude is S, find the lateral surface area of die prism. 

22. If S is the lateral surface area, a is the altitude, and p is the perimeter of 
the base of a right prism, prove that 

S = ap. 

23. Use the formula in Exercise 22 to find the lateral surface area of tiia 
prism in Exercise 21. Does your answer agree with the one obtained in 
Exercise 2 1 ? 

24. Find the altitude of a right prism if the lateral surface area is 336 and 
the perimeter of the base is 28. 

25. Find the perimeter of the base of a right prism if the lateral surface area 
is 351 and the altitude is 13^. 

26. If the base of the right prism in Fxercise 25 is a square region, find the 
length of each of its edges. 

27* Find the total surface area of a right triangular prism if the boundary 

of each base is an equilateral triangle whose sides have length 10 and If 

the altitude of the prism is 12. 
2S. Ilie area of a cross section of a prism is 32. The lateral surface area is 

128- Find the total surface area of the prism. 
29. If XB and FQ are two lateral edges of a prism, prove that AB and TQ 

arc eoplanar. 



672 Areas »r»d Volumes of Solids 



Ctiapler 15 



30. Prove Corollary 15.2 J, 

31 , Given the rectangular parol lelepiped shown in the figure with AB = 12, 
BC = 6, and CD = 8, find the length of die diagonal Ail, (Hint: Draw 
At?. What kind of triangle is A ABC? What kind of triangle is AACD?) 




32. In the figure RS is a diagonal of a cube and the length of each edge of 
the cube is 8. Prove that RS = 8^/5". 




33. Prove that the length of every diagonal of a tube is e y/S, where e is the 
length of one of its edges. 

34. challenge problem. Use the Distance Formula for a three-dimen- 
sional coordinate system to prove that the diagonals of a rectangular 
parallelepiped have equal Lengths. 

35. challenge PROBLEM. If h is the altitude of a prism, prove that h <, r, 
where r is the length of any one of its lateral edges. 



15.3 PYRAMIDS 

Figure 15-8 on page 673 shows some pictures of pyramids. Com- 
pare Figure 15-8 with Figure 15-1, In what respect does a pyramid 
differ from a prism? How arc they similar? 

Since a pyramid is similar in many respects to a prism , some terms 
for parts or' a prism are also used for parts of a pyramid. We shall use 
these terms without giving formal definitions. 



15.3 Pyramids 673 




Figure I5^i 



Definition 15.7 (See Figure 15-9,) Let Rka polygonal re- 
gion in a plane o and V a point not in a* For each point P of 
R there is a segment FV. The union of all such segments is 
called a pyramid. The polygonal region R is called the hase 
and V is called the vertex of the pyramid. The distance VT 
from V to a is the altitude of the pyramid. 




tx 



Figure 15*9 



For the pyramid in Figure 15-9, AV is a lateral edge and the tri- 
angular region ABV is a lateral face, Name four other lateral edges 
and four other lateral faces of the pyramid shown. How many edges 
lateral and base) does Ida pyramid ki\v m ,,11- IIow main laces in all? 
How many vertices in all? 

A cross section of a pyramid is the intersection of the pyramid with 
a plane parallel to the l*ase provided the intersection contains more 
than one point. 



674 Areas and Volumes of Solids 



Cfneiir 15 



THEOREM 15.4 The boundary of each cross s cction of a tria agu - 
tar pyramid is a triangle similar to the boundary of the base, and the 
areas of any two cross sections arc proportional to Lhe squares of 
the distances of their planes from the vertex of the pyramid. 

Proof: Tjet the triangular region ABC in plane a he the base of the 
pyramid as shown in Figure 15-10. LcL fi be a plane parallel to a and 
intersecting AV, BV, and EV in distinct points A\ W t and C. respec- 
tively. Then the triangular region A'B'C is a ctoss section gf the pyra- 
mid. Let S be the area of A ABC, let S' be the area of AA'B'C, let Jt be 
the distance from the vertex lo the cross section plane, and let h be the 
altitude of the prism. In Figure 1510, Jt = VF and k = VP. 



k = VP' 

h = VP 




Figure 15-10 

To complete the proof of the theorem we shall prove statements 
1 and 2. 

L AA'B'C ~ AABC 
2. (S\S) = (**,**) 

If/? = «,lhenA'=A.H' = B, C = C, Y = F t and k = h. There- 
fore AA'B'C - AABC and hence 

A A'B'C s AABC. 

It follows that AA'B'C' - AABC and that 

S' = S, k = h, and (S', S) = (&*, *■). 

Suppose, then, that £ ^ «. We have V 2 A, F A VF coplanar (Why?) 
with A-A'-V and F-F-V. AT ± Wand A*F i VF. Why? Therefore 



15.3 Pyramid* 675 

AA'FV- A APV by the A.A. Similarity Theorem. (lA'VF^lAVP 
LA'VV= ZAPV/i Hence 



(VA' t VA) = {k. h). 

In the same way, we can show that AB'FV — ABFV and hence 

(VB-, VB> = (*, /i). 
Therefore 

£VA' T VA) = (VB' t VB) 

and 

AA'W - AAVi* 

by the S. A, S. Similarity Theorem. Therefore 
(A'B',AB) = (VA\VA) 



and 



I A'B\ AS) = (ft, h) 
A'B' = 4'AJ3. 



In the same way it can be shown that 

ft 

h 



= £*8C 



and 

CA' = 

Then 



C'A' = ^>CA. 



{A'B f , B'C, CA') = (AB, BC, CA) 

and it follows that AA'B'C <- AABC by the S. S. S. Similarity Theo- 
rem. Recall that in Chapter 10 it was proved (Theorem 10.15) that if 
two triangles arc similar, then their areas are proportional to the square 
of the lengths of any two corresponding sides. Therefore 

(S\ S) s ((A'/?)*, (AB)*). 
Since 

(A'B 1 , AB) = (ft, h) t 

it follows that 

{(A'l?)\ (AB)*) = (ft*, V). 
Therefore 

(S\ S) = (ft». M)» 

and the proof is complete. 



676 ATM* and Volumes of Solids 



Chapter 15 



Compare our next theorem for pyramids with Theorem 15.2 for 
prisms. 

THEOREM 15.5 In any pyramid the areas of any two cross sec- 
lions arc proportional to the squares of the distances of their planes 
from die vertex of the pyramid. 

Proof: Let a pyramid be given as shown in Figure 15-1 1, (We have 
shown a pyramid whose base is a polygonal region consisting of five 
sides. The following argument can be modified to apply to a pyramid 
whose base is a polygonal region consisting of n sides, where n > 3.) 



k m VP' 
h=VP 




Figure 15-11 B 



Let the polygonal region ABCDE be the base and let A'WC&W 
be any cross section of the pyramid* Let S be the area of Ihe region 
ABCDE and let S' be the area of the region A'B'CD'E'. Let k be the 
distance from the vertex to the cross section plane and let h be die alti- 
tude of the pyramid. Hie region ABCDE can be divided into nonover- 
lapping triangular regions tu t 2 , h as shown in Figure 15-1 L Let t{, t' 2 , 
t$ be die corresponding triangular regions mA'E'CD'F/. Let Tt. 7 a, T s 
be the areas of t u r 2 , # y , respectively, and let If, Ti, 'ft be the areas of 
t{, <2» fii, respectively. 



Thci 



s- = n + n + r 3 



S = T, + T 2 + T; 



3- 



15.3 Pyramids 677 
By Theorem 15.4, 

From the product property of a proportion it follows that 

Tift* = TiR 



Similarly, 
and 



T 2 h* = T 2 fc* 

r s h* = raft?. 

Then adding and using the Distributive Property, we get 

pi + n + n)h £ = (Tt + r 3 + rajtf 

S'Ji* = Sfc* 
(S\ S) = (fc 2 , **) 
and this completes the proof. 

We now use Theorem 15.5 to prove our next theorem. 

THEOREM 15.6 (The Pyramid Crow Section Theorem) If two 
pyramids have equal altitudes and if their bases have equal areas, 
then cross sections equidistant from the vertices have equal areas. 

Proof: Let two pyramids be given as shown in Figure 15-12. (We have 
shown pyramids whose bases are triangular regions. The theorem, how- 
ever, is not restricted to this case and our proof applies equally as well 
to pyramids whose bases are polygonal regions with more than 3 sides.) 





Figure 1!S- 12 



Areas and Volumes of Solids 



Chapter 15 



Let the base area of each pyramid be S, let h be the altitude of each, and 
let k be the distance from the vertex to the plane of the cross section of 
each. Let the areas of the cross sections be Si and S2. We must prove 
that Sj = Sg. By Theorem 15.5, 



Therefore 



(Si. S) f (*t, A*) = (fig, S). 



Si = s 2 , 



and the proof is complete. 



EXERCISES 15.3 

In Exercises 1-13, copy and complete each statement. Exercises 3-13 refer 
Lo Ore pyramid shown in Figure 15-13 in which P is a point in a, the phi n ltd 
the base, such that f la, 




Figure 15- IS 



1. All the lateral faces of & pyramid are [7] regions. 

2. If the boundary of die base of a pyramid is an equilateral triangle, then 
the boundary of each cross section is |Tj. 

3. The region ABCD is called the ]T] of the pyramid. 

4. V is called the G] of the pyramid. 
5- A V is called a [?) of the pyramid* 

6, There are [?] lateral edges in all 

7. Counting the lateral edges and the edges of the bases, there are \T\ edges 
in all. 

8. The triangular region AAV is called a of the pyramid. 

9, There are [fj lateral faces in all. 



15.3 Pyramids 679 

10- Counting the base, the total number of faces is [TJ. 

11. There are \J} vertices in all. 

12. If Vis the number of vertices, K the number of edges, and F the number 
of faces, then V - E + P = |7J. 

13- VP is the [T] of the pyramid, 

14. Compare your answer to Exercise 12 with that of Exercise 14 in Exer- 
cises 15.2. Are they the same? 

15. Compute V - E + F(scc Exercise 12) for the pyramid shown in Figure 
15-11 and for the prism shown in Figure 15-5, Are your answers the 
same for both solids? Do you think V — E + F = 2 for every prism or 
pyramid? Try some more examples using figures from this chapter. 

16. Recall that the center of a regular polygon is the center of the circum- 
scribed circle. The center of a regular polygonal region is the center of 
the polygon hounding the region. A pyramid is a regular pyramid if and 
only if its base is a regular polygonal region and the foot of the perpen- 
dicular from its vertex to its base is the center of the base. Figure 15-13 
shows a regular pyramid whose base is a square. Prove that the bound- 
ary of the lateral face A VB is an isosceles triangle. (Hint: Draw PA and 
PBand prove that &AFV m ABPVby the S. A. S. Postulate.) 

17. Given that the pyramid shown in Figure 15-13 is a regular pyramid 
whose base is a square region (see Exercise 16), prove that 
AAVB m ABVC. 

18. Draw a picture of a regular pentagonal pyramid, that is, a regular pyra- 
mid whose base is a regular pentagonal region (see Exercise 16). Label 
the vertex Vand the vertices of the base A, B t C, D t E. Pick any two 
lateral faces and prove that they arc congruent isosceles triangles. (Hint: 
Draw the perpendicular from V to the center of the base as in Figure 
15-33.) 

19. Recall that corresponding altitudes of congruent triangles are con- 
gruent. Prove that the lateral surface area of the regular pyramid shown 
in Figure 15-13 is given by S = £ap, where p is the perimeter of the 
base and a is the length of the segment whose ondpoints are the vertex 
of the pyramid and die fool of the perpendicular from the vertex to an 
edge of the base, (See Exercises 16, 17, and 18.) 

20. Let two pyramids, one triangular and one hexagonal, with equal base 
areas be given. The altitude of each pyramid is 9 in. The cross section of 
the triangular pyramid that is 3 in. from the base has an area of 40 sq. in. 
What is the area or the cross section of the hexagonal pyramid that is 
3 in. from its lrase? 

21. The area of the base of a pentagonal pyramid is 1024. The distance from 
the vertex to the plane of a cross section is 3 and the altitude of the pyra- 
mid is 8. Find the area of the cross section. ; J Jink Use Theorem 15.5.) 



680 Art** and Volumes of Solids Chapter 15 

22, The boundary of the base of a pyramid is an equilateral triangle, and the 
boundary of each lateral face is an equilateral triangle with sides of 
length 8. Find the total surface area of the pyramid. 

23, The boundary of the base of a pyramid is a square whose sides are 
10 em. in length, and the boundary of each lateral face is an equilateral 
triangle. Find the total surface area of the pyramid. 

24, Find the altitude of the pyramid of Exercise 23, 

25, Find the altitude of the pyramid of Exercise 22. 

26, The area of a cross section of a pyramid is 125. The area of the base is 
405. The altitude of the pyramid is 9. Find the distance k from the vertex 
to the plane of the cross section. 

fr. Given (a, b) = (c, d) t prove that (a*, fe 2 ) = (c 2 , <#). If 

(<j, b) = (c, d) 

with constant of proportionality t> then what is the constant of pro- 
portionality for 

{«*, £*) = ( C 2, tP)? 

Refer tr> the proof of Theorem 15.4. In which sentence of this proof is 
this property of a proportionality used? 



15.4 AREAS AND VOLUMES OF PRISMS AND CYLINDERS 

As stated in the inrrixhn.-i inn i<> this chapter, we shall not t-cat areas 
and volumes of solids as rigorously as we did areas of polygonal regions. 
The concepts of surface area and volume are natural extensions of the 
area concept developed in Chapter 9. We accept without proof the 
fact that each solid discussed in this chapter has a surface area and a 
volume. 

For volumes, we accept two postulates and use them to prove the 
volume formulas for prisms, cylinders, pyramids, and cones. Later in 
the chapter we develop a formula for the volume of a sphere. 

For surface areas of cylinders, cones, and spheres, we develop the 
formulas informally and then state them formally as theorems whose 
proofs are beyond the scope of this book. 

Our first postulate is similar to Postulate 28, Lhe Rectangle Area 
Postulate of Chapter 9. 

POSTULATE 31 (Rectangular Parallelepiped Volume Postu- 
late) The volume of a rectangular parallelepiped is the product of the 
altitude and the area of the base. 



15.4 Areas and Volumt* of Prisms and Cylinders 681 



Figure 15-14 is a picture of a rectangular parallelepiped. By defini- 
tion, its base is a rectangular region and each of its faces is a rectangular 
region. Thus any one of its faces could be called the base, and the 
length of any one of its edges that is perpendicular to that face could be 
called the altitude. 




Hfur* 1544 

By Postulate 31, the volume Vof the rectangular parallelepiped shown 
in Figure 15-14 is given by the formula 

V=Sh, 

where S is the area of the base and h is the altitude. By Postulate 28, 
S = oh. Therefore we have 

V = abh 

as a formula for the volume of a rectangular parallelepiped. 

Suppose that an ordinary deck of playing cards is arranged so that 
its lateral faces are vertical as suggested in Figure 15-15a. It seems rea- 
sonable dial if the shape of the deck is changed as in Figure 15-15b, the 
volume of the deck remains the same. It also seems reasonable that the 
fifteenth card from the bottom in Figure 15- 15a has the same volume 
as the corresponding card in Figure 15-15b. 




Figure IS- IS 



682 Areas and Volumes of Solids Chapter 15 

As a second example,, suppose that we make an approximate model 
of a rectangular pyramid by forming a slack of thin cards as suggested 
in Figure 15-16. Of course, the thinner we make the cards, the more 
cards there will be and the closer the approximation will be. Suppose 
that the cards in die model are kept at the same level but are allowed 
to change position by sliding along each other. 








Figure 15-16 

Then the shape of the model changes, but its volume does not change. 
For different positions of the cards, the base area and the thickness 
of any two cards at the same level are the same, hence their volume 
is the same. The total volume of the model is the total volume of the 
cards, and the total volume does not change when the cards slide along 
each other. 

More generally, imagine two solids with equal altitudes and with 
bases in the same horizontal plane. Suppose that every two cross sec- 
tions of these solids at equal distances from the bases have equal areas. 
Then it seems reasonable that the two solids should have equal vol- 
umes. The reason is that if we imagine die solids being eut into thin 
slices by planes parallel to the bases, the volumes of slices at the same 
distance from the bases will be approximately equal. Therefore the 
volumes of the two solids should be equal. 

The principle that we have tried to make plausible here is called 
Gavalieri's Principle after Professor Bonaventum CavaKeri (1598- 
1647) of the University of Bologna. He used this principle m obtaining 
some results that we now find in the calculus. We state this principle, 
formally as the second postulate of this section. 

POSTULATE 32 (Cflw/iWi Principle) If two solids have equal 
altitudes, and if cross sections of these solids at equal distances from 
the bases have equal areas, then the solids have equal volumes. 



15.4 Area* and Volumes of Prisms and Cylinders 683 

Cavalicri's Principle is the key to calculating volumes other than 
rectangular parallelepipeds. We use the principle in the proof of our 
next theorem, 

THEOREM 15. 7 The volume V of any prism is the product of it? 
altitude h and the area S of its base, that is, V = Sfc. 



Let a prism with altitude h and base area S be given as shown 
in Figure 15-17a, Let the rectangular parallelepiped shown in Figure 
15-t7b have the same altitude h and the same base area S T and let its 
base and the base of the given prism be in the same plane. 




Figure 15*17 

By the Prism Cross Section Theorem, all cross sections for both prisms 
have the same area S. By CavalierTs Principle, the two prisms have the 
same volume. It follows from the Rectangular Parallelepiped Volume 
Postulate that V = Sh for the given prism. 

Example I The base boundary of a prism is an equilateral triangle 
8 in. on a side. Its altitude is 12 in. Find the volume of the prism. 

Solution: The volume of the prism is given by V = $h t where S is the 
area of the triangular base and h is the altitude. We have S = 16 \/3 
(show this) and h = 12. Therefore 

V = 16- V3*I2 = 192 V3, 

and the volume is 192 v3"cu. in. 



684 



Areas and Volumes of Solids 



Ch»pt«r 15 



Figure 15-18 shows a diagram of a circular cylinder. Many of the 
terms used in defining a prism apply equally as well to a circular cylin- 
der. The base of a prism is a polygonal region. 




Figure 15-1* 

The base of a circular cylinder is a circular region, that is, the union 
of a circle and its interior. Make appropriate changes in the wording 
of the definition of a prism (Definition 15.1), refer to Figure 15-18, and 
define a circular cylinder. 

There are other cylinders in addition to circular cylinders, that is, 
cylinders whose bases are not circular regions; but we shall consider 
only circular cylinders in this text. Therefore, when we speak of a 
cylinder, we mean a circular cylinder. 

If QQ' in Figure 15-18 is perpendicular to a, then the cylinder is 
called a right circular cylinder, The altitude, bases, and cross sections 
of a cylinder are defined in (he same way as are the corresponding 
parts of prisms. 

The following two theorems are analogous to Theorems 15.1 and 
15,2 for prisms and can be proved in a similar way. We omit the details 
of the proofs. 

T1IEOR FM 15.S Hi© bou udary of each cross section of a cyl i n der 
is a circle that is congruent to the boundary of the base. 



Outline of Proof: (See Figure 15-19 on page 685.) Let R be the lyase 
of the given cylinder and let H' be any cross section of the cylinder- 
Let C be the center of the circle that bounds the base, let P be a point 
on that circle, and let C and F be the corresponding points in R' t Let 



15.4 Areas «nd Volumes of Prisms and Cylinders 685 

CP = t and let CF = t>, Then PCCF is a parallelogram (Why?), and 

FC = r f = PC = r. 

Since PC has a constant value regardless of the position of P on the 
base circle, then FC has a constant value. Thus all points F lie on a 
circle with radius r* and center U. 




Figure 15-19 

Therefore the cross section is a circular region and its boundary is a 
circle of radius r. Then its boundary is congruent to the boundary 
of the base. 

THEOREM 15.9 (The Cylinder Cross Section Theorem) The 
area of a cross section of a cylinder is equal to the area of the base. 

Proof: Assigned as an exercise* 

Cavalieri's Principle is used in the proof of the following theorem on 
the volume of a cylinder. 

THEOREM 1 5 JO The volume of a cylinder is the product of the 
altitude and the area of the Imss. 

Proof: The proof is similar to that of Theorem 15.7 and is assigned as 
an exercise. 

Imagine slitting a right circular cylinder and unrolling its lateral 
surface onto a plane. Which figure is obtained? Figure 15-20 on page 
081 BBggesis that the boundary of the region khufl obtained k a sec- 
tangle whoso altitude is the altitude of the cylinder and whose base 
is the circumference of the base of the cylinder. Thus, if -r is the radius 



sac 



Areas, and Volumes of Solids 



Chapter 15 



of the boundary of the base of a right cylinder and h is the altitude, 
then the lateral surface area of the cylinder is equal to the area of the 
rectangular region obtained by "unrolling" the cylinder, that is, the 
lateral surface area is 2vrh, Since the area of each of the circular bases 

is tff 2 , the total surface arm of a right cylinder is 27rrh + 2irA 



-L 




Figure 1 5-3)0 

We state these results formally as our next theorem. 

THEOREM 15.11 The lateral surface area of a circular cylinder 
of base radius r and altitude h is 2irrh and its total surface area is 
2irrk + Swr 6 . 

liliniiflb 2 Find the total surface area of a cylinder if the radius of 
the circle that bounds the base is 7 and the altitude of the cylinder is 10. 

Solution; The lateral surface area is 

2irrh= 2ff-7« 10 = Mto 
The area of each base is 

at* = ??(7) 2 = 49tt. 
Therefore the total surface area is 

140*7 + 2-49* = 140^ + 9&r = 23&r. 



EXERCISES 15.4 

1, A rectangular tank 6 ft, by 4 ft. is used for watering horses. If the tank is 
filled with water to a depth of 3 ft., how many cubic feet of water are 
in the tank? 

2* One gallon of water occupies 231 cu. in, of space- To the nearest hun- 
dredth, how many gallons of walcr are contained in a space of 1 cu. ft.'? 

3. To the nearest gallon, how inany gallons of water are in the tank of 
Exercise 1? (Sec Exercise 2.) 

4, Show that the volume V of a cube of side e is given by V = *3. 



15,4 Ar»a* and VoJumw of Prhmt and Cylinders 687 

5. Find the volume of a cube whose edge is 6 in, Find its total surface area. 

6. Write a formula for the volume Vof a cylinder if its altitude is h and the 
radius of tts base circle is r. 

7. A water tank in the shape of a cylinder is 40 ft. high. The diameter of its 
base is 2S ft. Find the volume of the tank. 

8. To the nearest thousand gallons, how many gallons of water will the 
tank of Exercise 7 hold? (Use w = -^- and see Exercise 2.) 

9. The altitude of a cylinder is 8 in. and the diameter of its ba.se circle is 
3 in. Kind the volume and total surface area of the cylinder. 

10. On a shelf in Roy's supermarket there are two cylindrical cans of coffee. 
The first is 1 j times as tall as the second, but the second has a diameter 
1^ times that of the first. How should Roy price the second can in rela- 
tion to the first if he wants the price per unit of volume to be the same 
for both cans? 

11. How do the volumes of two cylinders compare if their altitudes are the 
same but the radius of the base circle of the second cylinder is three 
times that of the first? 

12. How do die volumes of two cylinders compare if the radii of their bass 
circles are the same but the altitude of the second cylinder is throe 
times that of the first? 

13. Draw a suitable figure and prove Theorem 15.9. 
14 I3raw a suitable figure and prove Theorem 15.10, 

15. A brick chimney In the form of a cylindrical shell and 2o ft. tall is to be 
built. The inside and outside diameters are 24 in. and 10 in., respec- 
tively. If it takes 31 bricks per cubic foot of chimney, find die approx- 
imate number of bricks needed (Use v = 3.14.) 

16. An air conditioning unit is to be installed in a rectangular building. In 
order to install the correct size unit, it is necessary to know the number 
of cubic feet of air inside the building. If the dimensions of the building 
are as shown in Figure 15-21 , find the volume of air inside the building. 




17. A block of wood in the shape of a cube has edges 16 in. in length. A 
circular hole 7 in. in diameter is bored through the block from top to 
bottom. Find die volume of the part of the block that remains. 

18. In Exercise 17, if e is the length of die edge of the cube and r is the 
radius of the circular hole, write ft formula for the volume Vof die block 
that remains after the hole has been bored. 



HI 



Areas and Volumes of Solids 



Chapter 15 



15.5 VOLUMES OF PYRAMIDS AND CONES 

Here wc develop formulas for calculating volumes of pyramids and 
cones and for the surface area of a cone. As in Section 15,4 , Cavalieri's 
Principle plays a key role in the proofs of theorems on volume. 

THEOREM 15,12 Two pyramids with the same altitude and the 
same base area have the same volume. 



Proof: Let two pyramids lws given as suggested in Figure 15-22. By 
the Pyramid Cross Section Theorem, corresponding cross sections of 
die two pyramids have the same area. Therefore* by Cavalier! s Prin- 
ciple, their volumes are the same. 







Figure 15-22 

Our next theorem provides a formula for calculating the volume of 
a triangular pyramid. Suppose that we are given a triangular pyramid 
with base ABC and vertex W. (See Figure 15-23a.) Next, wc take a 
triangular prism with die same base area and altitude as shown in 
Figure 15-23b, Imagine two planes cutting the prism and dividing it 
into three triangular pyramids as shown in Figure 15-24, (Name the 
two cutting planes that divide the prism as shown in Figure 15-24,) 

D. 

V — =»* 

w 





Figure 15-23 



'51 



(bj 



15,5 Volumu of Pyramids and Cones 689 






Figure 13-S4 

Pyramid (c) in Figure 15-24 has the same base area and the same alti- 
tude as the given pyramid in Figure 15-23, Therefore, by Theorem 
15.12, they have the same volume. In Figure 15»23b, D£\\AC, 
AD || C£, and ADEC is a parallelogram with AE one of its diagonals. 
Therefore A, f>, E, C are coplanar and AADE s A CCA. Why? Think 
of pyramids (a) and (b) in Figure 15-24 as having bases ADE and ECA, 
respectively, arid common vertex W. Then pyramids (a) and (b) have 
the same base area (Why?) and the same altitude (the distance from 
W lo plane A DEC). By Theorem 15.12. these two pyramids have the 
same volume. 

Next, consider WECB in Figure 15-23. We have WE \ EC, 
WB | UB. so WECB is a parallelograin with WC one of its diagonals. 
Therefore W t E t C t B are coplanar and A WEC a ACBW. Think of 
pyramids (b) and (c) in Figure 15-24 as having bases WEC and CBW, 
respectively, and common vertex A, Then pyramids (b) and (c) have 
the same base area and the same altitude and, by Theorem 15.12, these 
two pyramids have the same volume. Therefore all three pyramids, 
(a). (b) t (c) in Figure 15-24, have the same volume, say V, and the vol- 
ume of the prism in Figure 15-23 is 3 V. 

Xow consider ABC as the base of the prism in Figure 15-23. Let 
the area of AABC be S and let h be the altitude of the prism. Then 

3V = Sh and V = $Sh. 

But the given pyramid in Figure 15-23 has the same base area S and 
the same altitude h. Therefore the volume V of a triangular pyramid is 
given by the formula 

V= l 3 $h, 

where S is the area of its base and h is its altitude. This result is out next 
theort tii. 



690 Areas and Volumes of Solids 



Chapter 15 



THEOREM 15.13 The volume of a triangular pyramid is one- 
third the product of its base area and its altitude. 

The formula V = ^Sh holds for any pyramid as our next theorem 
states. 

THEOREM 15.14 The volume of a pyramid is one-third die 
product of its base area and its altitude. 



Let a pyramid with base area S and altitude h be given as 
shown on the left in Figure 15-25. Consider a triangular pyramid 
having the same base area t the same altitude, and with Its base and the 
base of the given pyramid in the same plane. It follows from the Pyra- 
mid Cross Section Theorem that cross sections of these two pyramids 
formed by the same plane have the same area. By Cavalieri's Principle, 
the two pyramids have the same volume. Since the volume of the tri- 
angular pyramid is yS/i, the volume of die given pyramid is also \Sh 
and the proof is complete. 




Figure 15-25 

Example 1 The dimensions of the base of a rectangular pyramid are 
7 cm, by 11 cm. and its altitude is 16 cm. Find its volume, 

Solution; The volume V* of the pyramid is given by the fonmila 
V m \Sh. where S is the area of the base and h is Ihe altitude. We have 



S = 77 and V = ^ • 77 • 10 = 410|. 
Therefore the volume is 4 10} cu. cm. 

Figure 15-26 shows a picture of a circular cone, Just as the defini- 
tion of a circular cylinder is analogous to the definition of a prism, the 
definition of a circular cone is analogous to the definition of a pyramid. 



15,5 Volume* of PynmWi and Cones 691 
V 




Figure 15-26 



Many of the terms used in defining a pyramid apply equally as well to a 
circular cone. The base of a pyramid is a polygonal region. The base of 
a circular cone is a circular region. Make appropriate changes in the 
wording of the definition of a pyramid (Definition 1 5.7), refer to Figure 
15-26, and define a circular cone. 

There are other cones in addition to circular cones, that is, cones 
whose bases are not circular regions; but wc consider only circular 
cones here. Therefore, when we speak of a cone, we mean a circular 
cone. Also, when we speak of the base circle of a cone, we mean the 
circle which is the boundary of the base. 



Definition 15.8 (See Figure 15-27.) If the center of the base 
circle of a cone is the foot of the perpendicular from the ver- 
tex V to plane a, the cone is called a right circular cone. 




Figure 1S-S7 



The following theorem is analogous to Theorems 15,4 and 15.5 on 
pyramids and can be proved in a similar way, We give only an outline 
of the proof and omit the details. 



692 



Af*« and Volumes of Solids 



Chapter 15 



THEOREM 15.15 (The Cone Cross Section Theorem) A cross 
section of a cone of altitude h, made by a plane at a distance k from 
the vertex, is a circular region whose area and the area r>f the base 
are proportional to fc 2 and h 2 < 

Outline of Proof: (See Figure 15-28,) Let R be the base of the given 
cone and let ft' be any cross section of the cone. Let C be the center of 
the base circle, let P be any point on that circle, and let C and P be the 
corresponding points in ft'. Ijet VA = h, VB = K CF = f, and 
CF^S. 



Figure 15-28 



2. 




Then 



AVU8~~ &VCA 


Why? 


(VC t k) = (VC t h) 


Why? 


AVCF-* AVCP 


Why? 


(VC^jlVCr) 


Why? 


(VC,k,/) = {VC,h,f) 


Why? 


(k,/) = (h,r), ft*. 


?{Kh), 



/ = 



T" 



3, Since PC has a constant value regardless of the position of the 
point F on the base circle, then FC has a constant value. Thus 
all points F lie on a circle with radius ¥■ and center C". The cor- 
responding circular region is the cross section. 

4. Let S' be the area of the circular cross section and let S be the 
area of the base. Then 

(S\ S) = [mV) 2 , **l f [{?)*, f*) = (*?, V) and (ff, S) = (**, k*). 
This completes the outline of the proof. 



15.5 Volumes of Pyramids and Cones 693 

Cavalicri's Principle is used in the proof of the following theorem 
which tells us how to find the volume of a cone, 

THEOREM J 5. J 6 The volume of a circular cone is one-third the 
product of die area of die base and the altitude. 

Proof: The proof is similar to that of Theorem 15.14 and is left as an 

exercise. 

Figure 15-29 shows a picture of a right circular cone. C is the center 
of the base circle and P is a point of that circle. We call the distance 
VP, which is the same for any point Pon the base circle, the slant height 
of the cone and denote it by s. 





Figure lS-2» 

If you imagine slitting the right cone of Fig^ire 15-29 along VP and 
unrolling its lateral surface onto a plane, you get the sector of a circle 
shown on the right in Figure 15-29, You learned in Chapter 14 that the 
area of a sector of a circle is one-half the product of the radius of the 
sector and the length of the intercepted arc. In the case of a sector ob- 
tained by "unrolling" a cone, the radius of the sector is the slant height 
of the cone, and the length of the intercepted arc is the circumference 
of the base circle of the cone. Therefore a formula for the lateral sur- 
face area of a right circular cone is $sC f where $ is the slant height of 
the cone and C is the circumference of the l»ase circle. If r Is the base 
radius, then 

%$C = 2$(2nr) 5= ms. 

If S is the total surface area of a right circular cone, s its slant height, 
and r the radius of its base circle, then a formula for the total surface 
area is 



S = J • * ■ 2<rrr + irr 2 or 
This brings us to our next theorem. 



S = <rrf(s + r). 



694 



Artat and Volumes of Solids 



Chapter 15 



THEOREM 15, 1 7 The lateral surface area of a right circular cone 
of slant height s, base radius r, and base circumference C is f»0, or 
ma, and its total surface area is «w(s + r)» 

Example 2 TTie slant height of a right circular cone is 12 and the 
radius of its base circle is 5. Find the lateral surface area and the total 
surface area of the cone. 

Solution: The lateral surface area is JsC, where 

s = 12 and C = 2irr = 2-n'B = lOir. 
Therefore the lateral surface area is 

^- 12 • l(fcr = 60sr. 

The total surface area S is given by the formula 

S = m{s + r). 



Therefore 



S = w*5(12 + 5} = 85?:, 



EXERCISES 15,5 

1. The length of one edge of the base of a regular triangular pyramid is 
12 in. and the altitude of the pyramid is 18 in. Find the lateral surface 
area of the pyramid. Find the total surface area. Find the volume. (Re- 
call that a regular pyramid is one whose base boundary is a regular poly- 
gon and that die eenter of the polygon is the foot of the perpendicular 
from the vertex of the pyramid,) 

Exercises 2-9 refer to the regular hexagonal pyramid in Figure 15-30, with 
VC = 1ft in., AB = 12 in., and P die projection of V on AB. 

2. Find CP, 

3. Find VP, 

4. Find the area of die lateral face A V7J, 

5. Find the lateral surface area 
of the pyramid. 

6. Find the area of A ABC. 

7. Find the area of the base 
of the pyramid. 

8. Find the total surface area 
of the pyramid. 

9. Find the volume of the pyramid. 

Figure 15-30 




15.5 Volumt* of Pyramid* and Cones 695 



10. A plane bisects the altitude of a pyramid and is parallel to the base of 
i he pyramid. Find the ratio of the volume of die pyramid above the 
bisecting plane to the volume of the given pyramid (Hint: To what 
numbers are the area of the cross section in the bisecting plane and the 
area of the base proportional?) 

11. Find the total surfoo© area and die volume of the right circular cone 
shown in the figure* 




12, Find the capacity in gallons of a right conical tank if it is 60 in* deep and 
if the radius of its circular top is 28 in- (To "see" the rank, invert the 
cone of Exercise 11. Use *r = * 7 *-.) 

13. In the figure a right circular cone stands inside a right circular cylinder 
of same base and altitude. Write a formula for the volume of that por- 
tion of the cylinder not occupied by the cone. 




14, Figure 15-31 shows two right cylinders having the same base area and 
the same altitude. Figure 15-3 lb shows two cones with a common ver- 
tex V inside the cylinder. If V is midway between the bases of the 
cylinder, show that the sum of the volumes of the two cones shown in 
Figure 13-31 b is equal to the volume of the cone shown in (a). 




(a) 




Figure 15-31 



696 



/ >■ ,i ,iiii 'j'n lurries ot Solids 



Chapter 15 



15. The volume of the cone shown in the figure is 250 en. in. and its alti- 
tude is 15 in. A second tone is cut from the first by a plane parallel to 
the base and 6 in. from the vertex. Find the volume of the smaller cone. 
(Hint: If S and S* are the areas of the bases of the larger and smaller 
cones, respectively, to what numbers are S r and S proportional?) 




15 in. 



Hi. challenge problem. If a plane parallel to the base of a cone (or 
pyramid) cuts off anodicr cone (or pyramid), then the solid between 
the cutting plane and the base is called a frustum. Figure 1 5-32b shows 
a frustum of a right eone. If the radius of the base circle is 9 in., the 
radius of the top circle 6 in., and the height of the frustum 12 in., find 
the volume of the fnistum. (Hint: First find x t where x is the altitude of 
the smaller cone in Figure 15-32a.) 




Figure 15-32 



(«) 




17. challenge problem. The figure (on page 697) shows a picture of a 
regular square pyramid, (Sec Exercise 1.) A plane parallel to the base 
intersects the altitude of the pyramid at a point whose distance from the 
vertex is one-third the distance from the vertex to the base. If the alti- 
tude of the pyramid is 24 in. and the length of an edge of the base is 
18 in., find the lateral surface urea and the volume of the frustum. (See 
Exercise 16.) 



15.6 Sphtrt* and Sph«rie*I R«glons 697 




15.6 SURFACE AREAS OF SPHERES AND VOLUMES OF 
SPHERICAL REGIONS 

The surface area of a sphere may be found very roughly by winding 
a string around a hemisphere and by covering with string a circular 
disk having the same radius as the sphere as suggested in Figure 15-33. 





Kifurc 15-33 

A comparison of the lengths of the two strings suggests that the surface 
area of the hemisphere is twice the area of the circle. Since the area of a 
circle is -jrr 2 , this suggests that the surface area of a hemisphere is 2irr a 
and that the surface area of a sphere is 4^. 

We now proceed to a more sophisticated approach for finding the 
surface area of a sphere, but first we need a definition. 



Definition 15.9 A spherical region is the union of a sphere 
and its interior. 



Areas and Volumes of Solids 



Chapter 15 



Think of slicing a spherical region into n thin slices of thickness t 
where nt is the diameter 2r of the sphere. (See Figure 15-34.) This par- 
titions the surface into n zones. 





Figure 15-34 

Figure 15-35 suggests a vertical cross section C of the sphere made 
by a plane through the center E of the sphere. The sphere has been 
sliced by equally spaced horizontal planes into n zones. 



Ji 



■ 

- ■ , 


1 r 


V'. 


,'-" R D 




Ts 




B7 




T; 



Figure 15-35 



One of these zones, Tg, for example, is the surface generated when 
the arc AB rotates about the line FE. Let B' be the point in which the 
tangent line to C at A intersects GB* Then AB is approximately equal 
to AB and the area of the surface generated when AB' rotates about 
FE is approximately equal to the area of the surface generated when 
AB rotates about 



15.6 Spheres and Spherical Regions 699 

We have AB' 1AE. Why? Let D be the point on HE' such that 
DA ± FK. Then AD X B'D. Why? Therefore &AB'D — AAEF. 
(Show this by finding two pairs of congruent corresponding angles in 
the two triangles.} Let s' = AB\ f a AD, r = AE, and x = AF. Then 

(/* f to *) 

and 

s'* = rt. 

Think of the zone T 2 as a narrow ribbon of width $> s is the length of the 

are AB and is approximately equal to s\ Then the area of the zone is 
approximately equal to the length of the ribbon, about 2mar, times the 
width of the ribbon & Therefore the area of the zone is approximately 
2ira»\ Now, 2tt.ts is approximately equal to 2ffxs'» and 

Znx$ f a 2vrt 

since s*i = rf . If we combine the areas of the n zones, we find the area 
S of the sphere is given approximately by 

S = n(2irrt) 
= <2*rr)(nr) 

= (2fTf)(2r) 

"Ine total error introduced by using areas of ribbons to approximate 
areas of zones can be made as small as desired if the thickness of the 
slices is made small enough. The formula 

S = 4wf* 

for die surface area of a sphere is an exact formula, Our approach has 
involved approximations, but our result is the correct one. In higher 
mathematics, the area of a sphere is carefully defined and the assertion 
that S = 4sr& is proved. We state it as a theorem, 

THEOREM 15.18 The surface area of a sphere is 4tt times the 
square of the radius of the sphere; that is, S = 4nr 2 . 

In our discussion concerning Figure 15-35* we showed that the area 
Z of a zone T of a sphere is approximately equal to 2irrt , where r is the 
radius of the sphere and t is the thickness of the zone. Actually, 

Z = 2ffrf 

is also an exact formula. 



700 



Areas and Volumn of Solids 



Chapter 15 



The formula for the volume of a sphere can now be obtained with- 
out difficulty. Xote that when we say "volume of a sphere/' we mean 
"volume of a spherical region." 

Suppose that the surface of the sphere is divided into n pieces of 
equal areas by 'latitude" circles and "longitude" circles, regular spaced 
as suggested in Figure 15-36. Suppose that the pieces arc denoted by 




figure Uk* 

s±. s 2 ,... t s„. For i = 1, 2 n f let P ( be a point of s t . That is, P T is a 

point of su P 2 is a point of *2, etc. Join each of the boundary points of 
Si to the center C of the sphere with a segment. The union of these 
segments and the piece $t encloses a portion of the spherical region 
which is approximately a pyramid whose altitude is the radius of the 
sphere and whose base is the piece s-j. (Note that we could "flatten" 
tine base by using a portion of the plane that is tangent to the sphere at 
P\. Tlie error introduced by this flattening of the base when accumu- 
lated for all n pieces st, can be made as small as desired by making n 
sufficientiy large,) 

Recall that die volume of a pyramid is ^Sh t where S is the area of 
the base and h is die altitude. Therefore die volume of one of the pyra- 
midal portions of the spherical region is approximately ]rS where r is 
the radius of the sphere and 5 is the area of each .% By summing the 
volumes of the portions of the sphere corresponding to all of the pieces 
sj» we find that the volume V of the sphere is given approximately by 

V = n(%rS) 
= (W(«5) 

= flrXV) 

= $*rr3. 



15.6 Spheres and Spherical Regions 701 

Although we used approximations to obtain the formula for the 
volume of a sphere, the formula 



V = 

is an exact formula. We write it formally as our last theorem. 

THEOB EM 15. 19 If r is the radius of a sphere, the volume of the 
sphere is jw 3 , 

EXERCISES 15.6 

In working the exercises in this set, do not use a replacement for sr un- 
less instructed to do so. 

1. Find the surface area and the volume of a sphere whose diameter is 12. 

2. A sphere having a diameter of 14 in. is placed in a cubical box. If the 
length of each edge of the box is 14 in., how many cubic inches of the 
volume of the box are not occupied by the sphere? (Use n = ^f-.) 

3. The radii of two spheres are 3 in. and 6 in., respectively. Find the ratio 
of their surface areas. Find the ratio of their volumes, 

4. One sphere has a diameter that is three times that of a second sphere 
Find the ratio of their surface areas. Find the ratio of their volumes. 

5. In Exercise 4, if the larger sphere has a volume of 38,808, what is the 
volume of the smaller sphere? Use v = If- and find the radius of each 
of the two spheres. 

6. Given a hemisphere, a cylinder, and a cone such that the radii of the 
base circles of the cylinder and the cone are equal to the radius of the 
hemisphere. If the cylinder and the cone have altitudes equal to the 
radius of the hemisphere, prove that the volume of the cylinder is equal 
to the sum of the volumes of the hemisphere and the cone. 

7. If the altitude of a circular cylinder is equal to the diameter of a sphere, 
and if the radius of the base circle of the cylinder is equal to the radius of 
the sphere, prove that the volume of the sphere is two- thirds the volume 
of the cylinder. 

8. A sphere with radius 3^ in. is divided into 14 equal sections by planes 
containing the same diameter of the sphere. Find the volume of each 
section. {Use w = -*jf%) 

9. What is the largest radius a sphere eon have if the numerical value of 

the surface area of the sphere is to be greater than or equal to the nu- 
merical value of the volume of the sphere? 
10. The volumes of a sphere and a circular cone are equal, and the radius of 
the sphere equals the ra